36853207 c programming ppt
TRANSCRIPT
Course Objectives
Be able to read and write C programs Understand all C language constructs Be able to use pointers Have a good overview of the Standard Library Be aware of some of C’s traps and pitfalls
Practical Exercises
Practical exercises are a very important part of the course
An opportunity to experience some of the traps first hand!
Solutions are provided, discuss these amongst yourselves and/or with the tutor
If you get stuck, ask If you can’t understand one of the solutions, ask If you have an alternative solution, say
Features of C
C can be thought of as a “high level assembler” Designed for maximum processor speed Safety a definite second! THE system programming language (Reasonably) portable Has a “write only” reputation
History of C
Developed by Brian Kernighan and Dennis Ritchie of AT&T Bell Labs in 1972
In 1983 the American National Standards Institute began the standardisation process
In 1989 the International Standards Organisation continued the standardisation process
In 1990 a standard was finalised, known simply as “Standard C”
Everything before this is known as “K&R C”
Standard C vs K&R C
Parameter type checking added Proper floating point support added Standard Library covered too Many “grey areas” addressed New features added
Standard C is now the choice All modern C compilers are Standard C The course discusses Standard C
A C Program
#include <stdio.h> /* comment */
int main(void){
printf("Hello\n");printf("Welcome to the Course!\n");
return 0;}
#include <stdio.h> /* comment */
int main(void){
printf("Hello\n");printf("Welcome to the Course!\n");
return 0;}
HelloWelcome to the Course!
HelloWelcome to the Course!
tells compiler about standard input and output functions (i.e. printf + others)
main function
“begin”
“end”
flag success to operating system
The Format of C
Statements are terminated with semicolons Indentation is ignored by the compiler C is case sensitive - all keywords and Standard
Library functions are lowercase Strings are placed in double quotes Newlines are handled via \n Programs are capable of flagging success or
error, those forgetting to do so have one or other chosen randomly!
Another Example
#include <stdio.h>
int main(void){
int a, b;
printf("Enter two numbers: ");scanf("%i %i", &a, &b);
printf("%i - %i = %i\n", a, b, a - b);
return 0;}
#include <stdio.h>
int main(void){
int a, b;
printf("Enter two numbers: ");scanf("%i %i", &a, &b);
printf("%i - %i = %i\n", a, b, a - b);
return 0;}
create two integer variables, “a” and “b”
read two integer numbers into “a”
and “b”
write “a”, “b” and “a-b” in the format specified
Enter two numbers: 21 1721 - 17 = 4
Enter two numbers: 21 1721 - 17 = 4
Variables
Variables must be declared before use immediately after “{”
Valid characters are letters, digits and “_” First character cannot be a digit 31 characters recognised for local variables
(more can be used, but are ignored) Some implementations recognise only 6
characters in global variables (and function names)!
Upper and lower case letters are distinct
printf and scanf
printf writes integer values to screen when %i is used
scanf reads integer values from the keyboard when %i is used
“&” VERY important with scanf (required to change the parameter, this will be investigated later) - absence will make program very ill
“&” not necessary with printf because current value of parameter is used
Integer Types in C
C supports different kinds of integers maxima and minima defined in “limits.h”
type format bytes minimum maximum
char %c 1 CHAR_MIN CHAR_MAX
signed char %c 1 SCHAR_MIN SCHAR_MAX
unsigned char %c 1 0 UCHAR_MAX
short [int] %hi 2 SHRT_MIN SHRT_MAX
unsigned short %hu 2 0 USHRT_MAX
int %i 2 or 4 INT_MIN INT_MAX
unsigned int %u 2 or 4 0 UINT_MAX
long [int] %li 4 LONG_MIN LONG_MAX
unsigned long %lu 4 0 ULONG_MAX
Integer Example
#include <stdio.h>#include <limits.h>
int main(void){
unsigned long big = ULONG_MAX;
printf("minimum int = %i, ", INT_MIN);printf("maximum int = %i\n", INT_MAX);printf("maximum unsigned = %u\n", UINT_MAX);printf("maximum long int = %li\n", LONG_MAX);printf("maximum unsigned long = %lu\n", big);
return 0;}
#include <stdio.h>#include <limits.h>
int main(void){
unsigned long big = ULONG_MAX;
printf("minimum int = %i, ", INT_MIN);printf("maximum int = %i\n", INT_MAX);printf("maximum unsigned = %u\n", UINT_MAX);printf("maximum long int = %li\n", LONG_MAX);printf("maximum unsigned long = %lu\n", big);
return 0;} minimum int = -32768, maximum int = 32767
maximum unsigned = 65535maximum long int = 2147483647maximum unsigned long = 4294967295
minimum int = -32768, maximum int = 32767maximum unsigned = 65535maximum long int = 2147483647maximum unsigned long = 4294967295
Character Example
#include <stdio.h>#include <limits.h>
int main(void){
char lower_a = 'a';char lower_m = 'm';
printf("minimum char = %i, ", CHAR_MIN);printf("maximum char = %i\n", CHAR_MAX);
printf("after '%c' comes '%c'\n", lower_a, lower_a + 1);printf("uppercase is '%c'\n", lower_m - 'a' + 'A');
return 0;}
#include <stdio.h>#include <limits.h>
int main(void){
char lower_a = 'a';char lower_m = 'm';
printf("minimum char = %i, ", CHAR_MIN);printf("maximum char = %i\n", CHAR_MAX);
printf("after '%c' comes '%c'\n", lower_a, lower_a + 1);printf("uppercase is '%c'\n", lower_m - 'a' + 'A');
return 0;} minimum char = 0, maximum char = 255
after 'a' comes 'b'uppercase is 'M'
minimum char = 0, maximum char = 255after 'a' comes 'b'uppercase is 'M'
Note: print integer value of character
Integers With Different Bases
It is possible to work in octal (base 8) and hexadecimal (base 16)
#include <stdio.h>
int main(void){
int dec = 20, oct = 020, hex = 0x20;
printf("dec=%d, oct=%d, hex=%d\n", dec, oct, hex);printf("dec=%d, oct=%o, hex=%x\n", dec, oct, hex);
return 0;}
#include <stdio.h>
int main(void){
int dec = 20, oct = 020, hex = 0x20;
printf("dec=%d, oct=%d, hex=%d\n", dec, oct, hex);printf("dec=%d, oct=%o, hex=%x\n", dec, oct, hex);
return 0;}
dec=20, oct=16, hex=32dec=20, oct=20, hex=20
dec=20, oct=16, hex=32dec=20, oct=20, hex=20
zero puts compiler into octal mode!
zero “x” puts compiler into hexadecimal
mode
Real Types In C
C supports different kinds of reals maxima and minima are defined in “float.h”
type format bytes minimum maximum
float %f %e %g 4 FLT_MIN FLT_MAX
double %lf %le %lg 8 DBL_MIN DBL_MAX
long double %Lf %Le %Lg 10 LDBL_MIN LDBL_MAX
Real Example
#include <stdio.h>#include <float.h>
int main(void){
double f = 3.1416, g = 1.2e-5, h = 5000000000.0;
printf("f=%lf\tg=%lf\th=%lf\n", f, g, h);printf("f=%le\tg=%le\th=%le\n", f, g, h);printf("f=%lg\tg=%lg\th=%lg\n", f, g, h);
printf("f=%7.2lf\tg=%.2le\th=%.4lg\n", f, g, h);
return 0;}
#include <stdio.h>#include <float.h>
int main(void){
double f = 3.1416, g = 1.2e-5, h = 5000000000.0;
printf("f=%lf\tg=%lf\th=%lf\n", f, g, h);printf("f=%le\tg=%le\th=%le\n", f, g, h);printf("f=%lg\tg=%lg\th=%lg\n", f, g, h);
printf("f=%7.2lf\tg=%.2le\th=%.4lg\n", f, g, h);
return 0;}
f=3.141600 g=0.000012 h=5000000000.000000f=3.141600e+00 g=1.200000e-05 h=5.000000e+09f=3.1416 g=1.2e-05 h=5e+09f= 3.14 g=1.20e-05 h=5e+09
f=3.141600 g=0.000012 h=5000000000.000000f=3.141600e+00 g=1.200000e-05 h=5.000000e+09f=3.1416 g=1.2e-05 h=5e+09f= 3.14 g=1.20e-05 h=5e+09
Constants
Constants have types in C Numbers containing “.” or “e” are double: 3.5,
1e-7, -1.29e15 For float constants append “F”: 3.5F, 1e-7F For long double constants append “L”: -
1.29e15L, 1e-7L Numbers without “.”, “e” or “F” are int, e.g.
10000, -35 (some compilers switch to long int if the constant would overflow int)
For long int constants append “L”, e.g. 9000000L
Warning!
#include <stdio.h>
int main(void){
double f = 5000000000.0;double g = 5000000000;
printf("f=%lf\n", f);printf("g=%lf\n", g);
return 0;}
#include <stdio.h>
int main(void){
double f = 5000000000.0;double g = 5000000000;
printf("f=%lf\n", f);printf("g=%lf\n", g);
return 0;}
f=5000000000.000000g=705032704.000000
f=5000000000.000000g=705032704.000000
double constant created because of “.”
constant is int or long but 2,147,483,647 is the
maximum!
OVERFLOW
Named Constants
Named constants may be created using const
#include <stdio.h>
int main(void){
const long double pi = 3.141592653590L;const int days_in_week = 7;const sunday = 0;
days_in_week = 5;
return 0;}
#include <stdio.h>
int main(void){
const long double pi = 3.141592653590L;const int days_in_week = 7;const sunday = 0;
days_in_week = 5;
return 0;}
creates an integer
constant
error!
Preprocessor Constants
Named constants may also be created using the Preprocessor– Needs to be in “search and replace” mode
– Historically these constants consist of capital letters
#include <stdio.h>
#define PI 3.141592653590L#define DAYS_IN_WEEK 7#define SUNDAY 0
int day = SUNDAY;long flag = USE_API;
#include <stdio.h>
#define PI 3.141592653590L#define DAYS_IN_WEEK 7#define SUNDAY 0
int day = SUNDAY;long flag = USE_API;
search for “PI”, replace with 3.1415....
“PI” is NOT substituted here
Note: no “=” and no “;”
Take Care With printf And scanf!
#include <stdio.h>
int main(void){
short a = 256, b = 10;
printf("Type a number: ");scanf("%c", &a);
printf("a = %hi, b = %f\n", a, b);
return 0;}
#include <stdio.h>
int main(void){
short a = 256, b = 10;
printf("Type a number: ");scanf("%c", &a);
printf("a = %hi, b = %f\n", a, b);
return 0;}
“%c” fills one byte of “a” which is two
bytes in size
“%f” expects 4 byte float in IEEE format, “b” is 2 bytes and
NOT in IEEE format
Type a number: 1a = 305 b = Floating support not loaded
Type a number: 1a = 305 b = Floating support not loaded
Summary
K&R C vs Standard C main, printf Variables Integer types Real types Constants Named constants Preprocessor constants Take care with printf and scanf
Operators in C
Arithmetic operators Cast operator Increment and Decrement Bitwise operators Comparison operators Assignment operators sizeof operator Conditional expression operator
Arithmetic Operators
C supports the arithmetic operators:
+ addition
- subtraction
* multiplication
/ division
% modulo (remainder)
“%” may not be used with reals
Using Arithmetic Operators
The compiler uses the types of the operands to determine how the calculation should be done
int main(void){
int i = 5, j = 4, k;double f = 5.0, g = 4.0, h;
k = i / j;h = f / g;h = i / j;
return 0;}
int main(void){
int i = 5, j = 4, k;double f = 5.0, g = 4.0, h;
k = i / j;h = f / g;h = i / j;
return 0;}
“i” and “j” are ints, integer division is done, 1 is
assigned to “k”
“f” and “g” are double, double division is done, 1.25
is assigned to “h”
integer division is still done, despite “h” being double. Value assigned is 1.00000
The Cast Operator
The cast operator temporarily changes the type of a variable
int main(void){
int i = 5, j = 4;double f;
f = (double)i / j;f = i / (double)j;f = (double)i / (double)j;f = (double)(i / j);
return 0;}
int main(void){
int i = 5, j = 4;double f;
f = (double)i / j;f = i / (double)j;f = (double)i / (double)j;f = (double)(i / j);
return 0;}
integer division is done here, the result, 1, is changed to a
double, 1.00000
if either operand is a double, the other is automatically
promoted
Increment and Decrement
C has two special operators for adding and subtracting one from a variable
++ increment
- - decrement
These may be either prefix (before the variable) or postfix (after the variable):
int i = 5, j = 4;
i++;--j;++i;
int i = 5, j = 4;
i++;--j;++i;
“i” becomes 6
“j” becomes 3
“i” becomes 7
Prefix and Postfix
The prefix and postfix versions are different
#include <stdio.h>
int main(void){
int i, j = 5;
i = ++j;printf("i=%d, j=%d\n", i, j);
j = 5;i = j++;printf("i=%d, j=%d\n", i, j);
return 0;}
#include <stdio.h>
int main(void){
int i, j = 5;
i = ++j;printf("i=%d, j=%d\n", i, j);
j = 5;i = j++;printf("i=%d, j=%d\n", i, j);
return 0;} i=6, j=6
i=5, j=6
i=6, j=6i=5, j=6
equivalent to: 1. j++; 2. i = j;
equivalent to: 1. i = j; 2. j++;
Truth in C
To understand C’s comparison operators (less than, greater than, etc.) and the logical operators (and, or, not) it is important to understand how C regards truth
There is no boolean data type in C, integers are used instead
The value of 0 (or 0.0) is false Any other value, 1, -1, 0.3, -20.8, is true
if(32)printf("this will always be printed\n");
if(0)printf("this will never be printed\n");
if(32)printf("this will always be printed\n");
if(0)printf("this will never be printed\n");
Comparison Operators
C supports the comparison operators:
< less than
<= less than or equal to
> greater than
>= greater than or equal to
== is equal to
!= is not equal to
These all give 1 (non zero value, i.e. true) when the comparison succeeds and 0 (i.e. false) when the comparison fails
Logical Operators
C supports the logical operators:
&& and
|| or
! not
These also give 1 (non zero value, i.e. true) when the condition succeeds and 0 (i.e. false) when the condition fails
int i, j = 10, k = 28;
i = ((j > 5) && (k < 100)) || (k > 24);
int i, j = 10, k = 28;
i = ((j > 5) && (k < 100)) || (k > 24);
Logical Operator Guarantees
C makes two important guarantees about the evaluation of conditions
Evaluation is left to right Evaluation is “short circuit”
if(i < 10 && a[i] > 0)printf("%i\n", a[i]);
if(i < 10 && a[i] > 0)printf("%i\n", a[i]);
“i < 10” is evaluated first, if false the whole statement is false (because false AND anything is false) thus “a[i] > 0”
would not be evaluated
Warning!
Remember to use parentheses with conditions, otherwise your program may not mean what you think
int i = 10;
if(!i == 5)printf("i is not equal to five\n");
elseprintf("i is equal to five\n");
int i = 10;
if(!i == 5)printf("i is not equal to five\n");
elseprintf("i is equal to five\n");
in this attempt to say “i not equal to five”, “!i” is evaluated first. As “i” is 10, i.e. non zero, i.e. true, “!i” must be false, i.e. zero. Zero is
compared with five
i is equal to fivei is equal to five
Bitwise Operators
C has the following bit operators which may only be applied to integer types:
& bitwise and
| bitwise inclusive or
^ bitwise exclusive or
~ one’s compliment
>> right shift
<< left shift
Bitwise Example
#include <stdio.h>
int main(void){
short a = 0x6eb9;short b = 0x5d27;unsigned short c = 7097;
printf("0x%x, ", a & b);printf("0x%x, ", a | b);printf("0x%x\n", a ^ b);
printf("%u, ", c << 2);printf("%u\n", c >> 1);
return 0;}
#include <stdio.h>
int main(void){
short a = 0x6eb9;short b = 0x5d27;unsigned short c = 7097;
printf("0x%x, ", a & b);printf("0x%x, ", a | b);printf("0x%x\n", a ^ b);
printf("%u, ", c << 2);printf("%u\n", c >> 1);
return 0;}
0x4c21, 0x7fbf, 0x339e28388, 3548
0x4c21, 0x7fbf, 0x339e28388, 3548
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x4c21 0100 1100 0010 0001
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x4c21 0100 1100 0010 0001
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x7fbf 0111 1111 1011 1111
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x7fbf 0111 1111 1011 1111
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x339e 0011 0011 1001 1110
0x6eb9 0110 1110 1011 10010x5d27 0101 1101 0010 01110x339e 0011 0011 1001 1110
7097 0001 1011 1011 100128388 0110 1110 1110 0100
7097 0001 1011 1011 100128388 0110 1110 1110 0100
7097 0001 1011 1011 1001 3548 0000 1101 1101 1100
7097 0001 1011 1011 1001 3548 0000 1101 1101 1100
Assignment
Assignment is more flexible than might first appear
An assigned value is always made available for subsequent use
int i, j, k, l, m, n;
i = j = k = l = m = n = 22;
printf("%i\n", j = 93);
int i, j, k, l, m, n;
i = j = k = l = m = n = 22;
printf("%i\n", j = 93);
“n = 22” happens first, this makes 22 available for assignment to “m”. Assigning 22 to “m” makes 22 available for assignment to “l” etc.
“j” is assigned 93, the 93 is then made available to printf for printing
Warning!
One of the most frequent mistakes is to confuse test for equality, “==”, with assignment, “=”
#include <stdio.h>
int main(void){
int i = 0;
if(i = 0)printf("i is equal to zero\n");
elseprintf("somehow i is not zero\n");
return 0;}
#include <stdio.h>
int main(void){
int i = 0;
if(i = 0)printf("i is equal to zero\n");
elseprintf("somehow i is not zero\n");
return 0;}
somehow i is not zerosomehow i is not zero
Other Assignment Operators
There is a family of assignment operators:
+= -= *= /= %=
&= |= ^=
<<= >>=
In each of these:
expression1 op= expression2
is equivalent to:
(expression1) = (expression1) op (expression2)
a += 27;a += 27;
a = a + 27;a = a + 27;
f /= 9.2;f /= 9.2;
f = f / 9.2;f = f / 9.2;
i *= j + 2;i *= j + 2;
i = i * (j + 2);i = i * (j + 2);
sizeof Operator
C has a mechanism for determining how many bytes a variable occupies
#include <stdio.h>
int main(void){
long big;
printf("\"big\" is %u bytes\n", sizeof(big));printf("a short is %u bytes\n", sizeof(short));printf("a double is %u bytes\n", sizeof double);
return 0;}
#include <stdio.h>
int main(void){
long big;
printf("\"big\" is %u bytes\n", sizeof(big));printf("a short is %u bytes\n", sizeof(short));printf("a double is %u bytes\n", sizeof double);
return 0;}
"big" is 4 bytesa short is 2 bytesa double is 8 bytes
"big" is 4 bytesa short is 2 bytesa double is 8 bytes
Conditional Expression Operator
The conditional expression operator provides an in-line if/then/else
If the first expression is true, the second is evaluated
If the first expression is false, the third is evaluated
int i, j = 100, k = -1;
i = (j > k) ? j : k;
int i, j = 100, k = -1;
i = (j > k) ? j : k;
int i, j = 100, k = -1;
i = (j < k) ? j : k;
int i, j = 100, k = -1;
i = (j < k) ? j : k;
if(j > k)i = j;
elsei = k;
if(j > k)i = j;
elsei = k;
if(j < k)i = j;
elsei = k;
if(j < k)i = j;
elsei = k;
Precedence of Operators
C treats operators with different importance, known as precedence
There are 15 levels In general, the unary operators have higher
precedence than binary operators Parentheses can always be used to improve
clarity#include <stdio.h>
int main(void){
int j = 3 * 4 + 48 / 7;
printf("j = %i\n", j);
return 0;}
#include <stdio.h>
int main(void){
int j = 3 * 4 + 48 / 7;
printf("j = %i\n", j);
return 0;}
j = 18j = 18
Associativity of Operators
For two operators of equal precedence (i.e. same importance) a second rule, “associativity”, is used
Associativity is either “left to right” (left operator first) or “right to left” (right operator first)
#include <stdio.h>
int main(void){
int i = 6 * 4 / 7;
printf("i = %d\n", i);
return 0;}
#include <stdio.h>
int main(void){
int i = 6 * 4 / 7;
printf("i = %d\n", i);
return 0;}
i = 3i = 3
Precedence/Associativity Table
Operator
Associativity() [] -> .
left to right
! ~ ++ -- - + (cast) * & sizeof
right to left
* / %
left to right
+ -
left to right
<< >>
left to right
< <= >= >
left to right
== !=
left to right
&
left to right
|
left to right
^
left to right
&&
left to right
||
left to right
?:
right to left
= += -= *= /= %= etc
right to left
,
left to right
Review
#include <stdio.h>
int main(void){
int i = 0, j, k = 7, m = 5, n;
j = m += 2;printf("j = %d\n", j);
j = k++ > 7;printf("j = %d\n", j);
j = i == 0 & k;printf("j = %d\n", j);
n = !i > k >> 2;printf("n = %d\n", n);
return 0;}
#include <stdio.h>
int main(void){
int i = 0, j, k = 7, m = 5, n;
j = m += 2;printf("j = %d\n", j);
j = k++ > 7;printf("j = %d\n", j);
j = i == 0 & k;printf("j = %d\n", j);
n = !i > k >> 2;printf("n = %d\n", n);
return 0;}
Control Flow
Decisions - if then else More decisions - switch Loops - while, do while, for Keyword break Keyword continue
Decisions - if then
Parentheses surround the test One statement becomes the “then part” If more are required, braces must be used
scanf("%i", &i);
if(i > 0)printf("a positive number was entered\n");
scanf("%i", &i);
if(i > 0)printf("a positive number was entered\n");
if(i < 0) {printf("a negative number was entered\n");i = -i;
}
if(i < 0) {printf("a negative number was entered\n");i = -i;
}
Warning!
A semicolon after the condition forms a “do nothing” statement
printf("input an integer: ");scanf("%i", &j);
if(j > 0);printf("a positive number was entered\n");
printf("input an integer: ");scanf("%i", &j);
if(j > 0);printf("a positive number was entered\n");
input an integer: -6
a positive number was entered
input an integer: -6
a positive number was entered
if then else
An optional else may be added One statement by default, if more are required,
braces must be used
if(i > 0)printf("i is positive\n");
elseprintf("i is negative\n");
if(i > 0)printf("i is positive\n");
elseprintf("i is negative\n");
if(i > 0)printf("i is positive\n");
else {printf("i is negative\n");i = -i;
}
if(i > 0)printf("i is positive\n");
else {printf("i is negative\n");i = -i;
}
Nesting ifs
else associates with the nearest if
int i = 100;
if(i > 0)if(i > 1000)
printf("i is big\n");else
printf("i is reasonable\n");
int i = 100;
if(i > 0)if(i > 1000)
printf("i is big\n");else
printf("i is reasonable\n"); i is reasonablei is reasonable
int i = -20;
if(i > 0) {if(i > 1000)
printf("i is big\n");} else
printf("i is negative\n");
int i = -20;
if(i > 0) {if(i > 1000)
printf("i is big\n");} else
printf("i is negative\n");i is negativei is negative
switch
C supports a switch for multi-way decision making
switch(c) { case 'a': case 'A':
printf("area = %.2f\n", r * r * pi);break;
case 'c': case 'C':printf("circumference = %.2f\n", 2 * r * pi);break;
case 'q':printf("quit option chosen\n");break;
default:printf("unknown option chosen\n");break;
}
switch(c) { case 'a': case 'A':
printf("area = %.2f\n", r * r * pi);break;
case 'c': case 'C':printf("circumference = %.2f\n", 2 * r * pi);break;
case 'q':printf("quit option chosen\n");break;
default:printf("unknown option chosen\n");break;
}
More About switch
Only integral constants may be tested If no condition matches, the default is executed If no default, nothing is done (not an error) The break is important
float f;
switch(f) { case 2:
....
float f;
switch(f) { case 2:
....
switch(i) { case 2 * j:
....
switch(i) { case 2 * j:
....
i = 3;
switch(i) { case 3: printf("i = 3\n"); case 2: printf("i = 2\n"); case 1: printf("i = 1\n");}
i = 3;
switch(i) { case 3: printf("i = 3\n"); case 2: printf("i = 2\n"); case 1: printf("i = 1\n");}
i = 3i = 2i = 1
i = 3i = 2i = 1
A switch Example
printf("On the ");switch(i) { case 1: printf("1st"); break; case 2: printf("2nd"); break; case 3: printf("3rd"); break; default: printf("%ith", i); break;}printf(" day of Christmas my true love sent to me ");switch(i) { case 12: printf("twelve lords a leaping, "); case 11: printf("eleven ladies dancing, "); case 10: printf("ten pipers piping, "); case 9: printf("nine drummers drumming, "); case 8: printf("eight maids a milking, "); case 7: printf("seven swans a swimming, "); case 6: printf("six geese a laying, "); case 5: printf("five gold rings, "); case 4: printf("four calling birds, "); case 3: printf("three French hens, "); case 2: printf("two turtle doves and "); case 1: printf("a partridge in a pear tree\n");}
printf("On the ");switch(i) { case 1: printf("1st"); break; case 2: printf("2nd"); break; case 3: printf("3rd"); break; default: printf("%ith", i); break;}printf(" day of Christmas my true love sent to me ");switch(i) { case 12: printf("twelve lords a leaping, "); case 11: printf("eleven ladies dancing, "); case 10: printf("ten pipers piping, "); case 9: printf("nine drummers drumming, "); case 8: printf("eight maids a milking, "); case 7: printf("seven swans a swimming, "); case 6: printf("six geese a laying, "); case 5: printf("five gold rings, "); case 4: printf("four calling birds, "); case 3: printf("three French hens, "); case 2: printf("two turtle doves and "); case 1: printf("a partridge in a pear tree\n");}
while Loop
The simplest C loop is the while Parentheses must surround the condition One statement forms the body of the loop Braces must be added if more statements are to
be executed
int j = 5;
while(j > 0)printf("j = %i\n", j--);
int j = 5;
while(j > 0)printf("j = %i\n", j--);
j = 5j = 4j = 3j = 2j = 1
j = 5j = 4j = 3j = 2j = 1while(j > 0) {
printf("j = %i\n", j);j--;
}
while(j > 0) {printf("j = %i\n", j);j--;
}
(Another) Semicolon Warning!
A semicolon placed after the condition forms a body that does nothing
int j = 5;
while(j > 0);printf("j = %i\n", j--);
int j = 5;
while(j > 0);printf("j = %i\n", j--);
program disappears into an infinite loop
• Sometimes an empty loop body is required
int c, j;
while(scanf("%i", &j) != 1)while((c = getchar()) != '\n')
;
int c, j;
while(scanf("%i", &j) != 1)while((c = getchar()) != '\n')
;
placing semicolon on the line below
makes the intention obvious
while, Not Until!
Remember to get the condition the right way around!
int j = 5;
printf("start\n");while(j == 0)
printf("j = %i\n", j--);printf("end\n");
int j = 5;
printf("start\n");while(j == 0)
printf("j = %i\n", j--);printf("end\n");
user probably intends “until j is
equal to zero”, however this is NOT
the way to write itstartend
startend
do while
do while guarantees execution at least once
int j = 5;
printf("start\n");do
printf("j = %i\n", j--);while(j > 0);printf("stop\n");
int j = 5;
printf("start\n");do
printf("j = %i\n", j--);while(j > 0);printf("stop\n");
startj = 5j = 4j = 3j = 2j = 1stop
startj = 5j = 4j = 3j = 2j = 1stop
int j = -10;
printf("start\n");do {
printf("j = %i\n", j);j--;
} while(j > 0);printf("stop\n");
int j = -10;
printf("start\n");do {
printf("j = %i\n", j);j--;
} while(j > 0);printf("stop\n");
startj = -10stop
startj = -10stop
for Loop
for encapsulates the essential elements of a loop into one statement
for(initial-part; while-condition; update-part)body;
for(initial-part; while-condition; update-part)body;
int j;
for(j = 5; j > 0; j--)printf("j = %i\n", j);
int j;
for(j = 5; j > 0; j--)printf("j = %i\n", j);
j = 5j = 4j = 3j = 2j = 1
j = 5j = 4j = 3j = 2j = 1
for(j = 5; j > 0; j--) {printf("j = %i ", j);printf("%s\n", ((j%2)==0)?"even":"odd");
}
for(j = 5; j > 0; j--) {printf("j = %i ", j);printf("%s\n", ((j%2)==0)?"even":"odd");
}
j = 5 oddj = 4 evenj = 3 oddj = 2 evenj = 1 odd
j = 5 oddj = 4 evenj = 3 oddj = 2 evenj = 1 odd
for Is Not Until Either!
Remember to get the for condition the right way around (it is really a while condition)
int j;
printf("start\n");for(j = 5; j == 0; j--)
printf("j = %i\n", j);printf("end\n");
int j;
printf("start\n");for(j = 5; j == 0; j--)
printf("j = %i\n", j);printf("end\n");
user probably intends “until j is
equal to zero”, however this is NOT
the way to write it either!
startend
startend
Stepping With for
Unlike some languages, the for loop is not restricted to stepping up or down by 1
#include <math.h>
int main(void){
double angle;
for(angle = 0.0; angle < 3.14159; angle += 0.2)printf("sine of %.1lf is %.2lf\n",
angle, sin(angle));
return 0;}
#include <math.h>
int main(void){
double angle;
for(angle = 0.0; angle < 3.14159; angle += 0.2)printf("sine of %.1lf is %.2lf\n",
angle, sin(angle));
return 0;}
Extending the for Loop
The initial and update parts may contain multiple comma separated statements
int i, j, k;
for(i = 0, j = 5, k = -1; i < 10; i++, j++, k--)
int i, j, k;
for(i = 0, j = 5, k = -1; i < 10; i++, j++, k--)
The initial, condition and update parts may contain no statements at all!
for(; i < 10; i++, j++, k--)for(; i < 10; i++, j++, k--)
for(;i < 10;)for(;i < 10;)
for(;;)for(;;)
use of a while loop would be clearer here!
creates an infinite loop
break
The break keyword forces immediate exit from the nearest enclosing loop
Use in moderation!
for(;;) {printf("type an int: ");if(scanf("%i", &j) == 1)
break;while((c = getchar()) != '\n')
;}printf("j = %i\n", j);
for(;;) {printf("type an int: ");if(scanf("%i", &j) == 1)
break;while((c = getchar()) != '\n')
;}printf("j = %i\n", j); type an int: an int
type an int: notype an int: 16j = 16
type an int: an inttype an int: notype an int: 16j = 16
if scanf returns 1, jump out of the loop
continue
The continue keyword forces the next iteration of the nearest enclosing loop
Use in moderation!
for(j = 1; j <= 10; j++) {if(j % 3 == 0)
continue;printf("j = %i\n", j);
}
for(j = 1; j <= 10; j++) {if(j % 3 == 0)
continue;printf("j = %i\n", j);
}
j = 1j = 2j = 4j = 5j = 7j = 8j = 10
j = 1j = 2j = 4j = 5j = 7j = 8j = 10
if j is exactly divisible by 3, skip
Summary
if (then) else - watch the semicolons switch can test integer values while, do while, for - watch the semicolons
again break continue
Functions
Rules of functions Examples - writing a function, calling a function Function prototypes Visibility Call by value The stack auto, static and register
The Rules
A function may accept as many parameters as it needs, or no parameters (like main)
A function may return either one or no values Variables declared inside a function are only
available to that function, unless explicitly passed to another function
Writing a Function - Example
int print_table(double start, double end, double step){
double d;int lines = 1;
printf("Celsius\tFarenheit\n");for(d = start; d <= end; d += step, lines++)
printf("%.1lf\t%.1lf\n", d, d * 1.8 + 32);
return lines;}
int print_table(double start, double end, double step){
double d;int lines = 1;
printf("Celsius\tFarenheit\n");for(d = start; d <= end; d += step, lines++)
printf("%.1lf\t%.1lf\n", d, d * 1.8 + 32);
return lines;}
accept 3 doubles when called
this is the TYPE of the value handed back
this is the ACTUAL value handed back
Calling a Function - Example
#include <stdio.h>
int print_table(double, double, double);
int main(void){
int how_many;double end = 100.0;
how_many = print_table(1.0, end, 3);print_table(end, 200, 15);
return 0;}
#include <stdio.h>
int print_table(double, double, double);
int main(void){
int how_many;double end = 100.0;
how_many = print_table(1.0, end, 3);print_table(end, 200, 15);
return 0;}
IMPORTANT: this tells the compiler how print_table works
the compiler knows these should be doubles and
converts them automatically
here the function’s return value is ignored - this is ok, if you don’t want it, you don’t have to use it
Calling a Function - Disaster!
#include <stdio.h>
int main(void){
int how_many;double end = 100.0;
how_many = print_table(1.0, end, 3);print_table(end, 200, 15);
return 0;}
#include <stdio.h>
int main(void){
int how_many;double end = 100.0;
how_many = print_table(1.0, end, 3);print_table(end, 200, 15);
return 0;}
now the compiler does not know how the function works
the compiler does NOT convert these ints to
doubles. The function picks up doubles
anyway!
Prototypes
The (optional) line
int print_table(double, double, double);
is known as a prototype If the compiler meets a call to an unknown
function it “guesses”– Guess 1: the function returns an int, even if it doesn’t
– Guess 2: you have passed the correct number of parameters and made sure they are all of the correct type, even if you haven’t
The prototype provides the compiler with important information about the return type and parameters
Prototyping is Not Optional
To achieve working programs the compiler is best given a prototype for each function called
When calling a Standard Library function, #include the file specified in the help page(s) - this file will contain the prototype
When calling one of your own functions, write a prototype by hand
Writing Prototypes
Prototype:
int print_table(double, double, double);int print_table(double, double, double);
int print_table(double start, double end, double step){
int print_table(double start, double end, double step){
Function header:
The function prototype may optionally include variable names (which are ignored)
int print_table(double start, double end, double step);int print_table(double start, double end, double step);
int print_table(double x, double y, double z);int print_table(double x, double y, double z);
Take Care With Semicolons
The prototype has a semicolon
The function header has an open brace
Don’t confuse the compiler by adding a semicolon into the function header!
int print_table(double start, double end, double step){
int print_table(double start, double end, double step){
int print_table(double start, double end, double step);{
int print_table(double start, double end, double step);{
int print_table(double start, double end, double step);int print_table(double start, double end, double step);
Example Prototypes
/* no parameters, int return value */int get_integer(void);
/* no parameters, double return value */double get_double(void);
/* no parameters, no return value */void clear_screen(void);
/* three int parameters, int return value */int day_of_year(int day, int month, int year);
/* three int parameters, long int return value */long day_since_1_jan_1970(int, int, int);
/* parameter checking DISABLED, double return value */double k_and_r_function();
/* short int parameter, (default) int return value */transfer(short int s);
/* no parameters, int return value */int get_integer(void);
/* no parameters, double return value */double get_double(void);
/* no parameters, no return value */void clear_screen(void);
/* three int parameters, int return value */int day_of_year(int day, int month, int year);
/* three int parameters, long int return value */long day_since_1_jan_1970(int, int, int);
/* parameter checking DISABLED, double return value */double k_and_r_function();
/* short int parameter, (default) int return value */transfer(short int s);
Example Calls
int i;double d;long l;short int s = 5;
i = get_integer();
d = get_double();
clear_screen();
i = day_of_year(16, 7, 1969);
l = day_since_1_jan_1970(1, 4, 1983);
d = k_and_r_function();d = k_and_r_function(19.7);d = k_and_r_function("hello world");
i = transfer(s);
int i;double d;long l;short int s = 5;
i = get_integer();
d = get_double();
clear_screen();
i = day_of_year(16, 7, 1969);
l = day_since_1_jan_1970(1, 4, 1983);
d = k_and_r_function();d = k_and_r_function(19.7);d = k_and_r_function("hello world");
i = transfer(s);
the compiler cannot tell which of these (if any) is correct - neither can we
without resorting to documentation!
no mention of “void” when calling these
functions
Rules of Visibility
C is a block structured language, variables may only be used in functions declaring them
int main(void){
int i = 5, j, k = 2;float f = 2.8F, g;
d = 3.7;}
void func(int v){
double d, e = 0.0, f;
i++; g--;f = 0.0;
}
int main(void){
int i = 5, j, k = 2;float f = 2.8F, g;
d = 3.7;}
void func(int v){
double d, e = 0.0, f;
i++; g--;f = 0.0;
}func’s “f” is used,
not main’s
compiler does not know about “d”
“i” and “g” not available here
Call by Value
When a function is called the parameters are copied - “call by value”
The function is unable to change any variable passed as a parameter
In the next chapter pointers are discussed which allow “call by reference”
We have already had a sneak preview of this mechanism with scanf
Call by Value - Example
#include <stdio.h>
void change(int v);
int main(void){
int var = 5;
change(var);
printf("main: var = %i\n", var);
return 0;}
void change(int v){
v *= 100;printf("change: v = %i\n", v);
}
#include <stdio.h>
void change(int v);
int main(void){
int var = 5;
change(var);
printf("main: var = %i\n", var);
return 0;}
void change(int v){
v *= 100;printf("change: v = %i\n", v);
}change: v = 500main: var = 5
change: v = 500main: var = 5
the function was not able to alter “var”
the function is able to alter “v”
C and the Stack
C uses a stack to store local variables (i.e. those declared in functions), it is also used when passing parameters to functions
The calling function pushes the parameters The function is called The called function picks up the parameters The called function pushes its local variables When finished, the called function pops its local
variables and jumps back to the calling function The calling function pops the parameters The return value is handled
Stack Example
#include <stdio.h>
double power(int, int);
int main(void){
int x = 2;double d;
d = power(x, 5);printf("%lf\n", d);
return 0;}
double power(int n, int p){
double result = n;
while(--p > 0)result *= n;
return result;}
#include <stdio.h>
double power(int, int);
int main(void){
int x = 2;double d;
d = power(x, 5);printf("%lf\n", d);
return 0;}
double power(int n, int p){
double result = n;
while(--p > 0)result *= n;
return result;}
main: x2
main: d?
power: p5
power: n2
power: result32.0
Storage
C stores local variables on the stack Global variables may be declared. These are not
stack based, but are placed in the data segment Special keywords exist to specify where local
variables are stored:
auto - place on the stack (default)
static - place in the data segment
register - place in a CPU register
Data may also be placed on the heap, this will be discussed in a later chapter
auto
Local variables are automatically allocated on entry into, and automatically deallocated on exit from, a function
These variables are therefore called “automatic” Initial value: random Initialisation: recommended
int table(void){
int lines = 13;auto int columns;
int table(void){
int lines = 13;auto int columns;
auto keyword redundant
static
The static keyword instructs the compiler to place a variable into the data segment
The data segment is permanent (static) A value left in a static in one call to a function
will still be there at the next call Initial value: 0 Initialisation: unnecessary if you like zeros
int running_total(void){
static int rows;
rows++;
int running_total(void){
static int rows;
rows++;
permanently allocated, but local to this
function
register
The register keyword tells the compiler to place a variable into a CPU register (you cannot specify which)
If a register is unavailable the request will be ignored
Largely redundant with optimising compilers Initial value: random Initialisation: recommended
void speedy_function(void){
register int i;
for(i = 0; i < 10000; i++)
void speedy_function(void){
register int i;
for(i = 0; i < 10000; i++)
Global Variables
Global variables are created by placing the declaration outside all functions
They are placed in the data segment Initial value: 0 Initialisation: unnecessary if you like zeros
#include <stdio.h>
double d;
int main(void){
int i;
return 0;}
#include <stdio.h>
double d;
int main(void){
int i;
return 0;}
variable “d” is global and available to all functions defined
below it
Review
Writing and calling functions The need for function prototypes Visibility C is “call by value” Local variables are stack based, this can be
changed with the static and register keywords
Global variables may be created, they are stored in the data segment
Pointers
Declaring pointers The “&” operator The “*” operator Initialising pointers Type mismatches Call by reference Pointers to pointers
Pointers - Why?
Using pointers allows us to:– Achieve call by reference (i.e. write functions which change
their parameters)
– Handle arrays efficiently
– Handle structures (records) efficiently
– Create linked lists, trees, graphs etc.
– Put data onto the heap
– Create tables of functions for handling Windows events, signals etc.
Already been using pointers with scanf Care must be taken when using pointers since
there are no safety features
Declaring Pointers
Pointers are declared by using “*” Declare an integer:
int i;int i;
Declare a pointer to an integer:
int *p;int *p;
There is some debate as to the best position of the “*”
int* p;int* p;
Example Pointer Declarations
int *pi; /* pi is a pointer to an int */
long int *p; /* p is a pointer to a long int */
float* pf; /* pf is a pointer to a float */
char c, d, *pc; /* c and d are a char pc is a pointer to char */
double* pd, e, f; /* pd is pointer to a double e and f are double */
char* start; /* start is a pointer to a char */
char* end; /* end is a pointer to a char */
int *pi; /* pi is a pointer to an int */
long int *p; /* p is a pointer to a long int */
float* pf; /* pf is a pointer to a float */
char c, d, *pc; /* c and d are a char pc is a pointer to char */
double* pd, e, f; /* pd is pointer to a double e and f are double */
char* start; /* start is a pointer to a char */
char* end; /* end is a pointer to a char */
The “&” Operator
The “&”, “address of” operator, generates the address of a variable
All variables have addresses except register variables
char g = 'z';
int main(void){
char c = 'a';char *p;
p = &c;p = &g;
return 0;}
char g = 'z';
int main(void){
char c = 'a';char *p;
p = &c;p = &g;
return 0;}
p c
'a'0x11320x1132
p g
'z'0x91A20x91A2
Rules
Pointers may only point to variables of the same type as the pointer has been declared to point to
A pointer to an int may only point to an int– not to char, short int or long int, certainly not to float,
double or long double
A pointer to a double may only point to a double– not to float or long double, certainly not to char or any of
the integers
Etc......
int *p; /* p is a pointer to an int */long large = 27L; /* large is a long int,
initialised with 27 */
p = &large; /* ERROR */
int *p; /* p is a pointer to an int */long large = 27L; /* large is a long int,
initialised with 27 */
p = &large; /* ERROR */
The “*” Operator
The “*”, “points to” operator, finds the value at the end of a pointer
#include <stdio.h>
char g = 'z';
int main(void){
char c = 'a';char *p;
p = &c;printf("%c\n", *p);
p = &g;printf("%c\n", *p);
return 0;}
#include <stdio.h>
char g = 'z';
int main(void){
char c = 'a';char *p;
p = &c;printf("%c\n", *p);
p = &g;printf("%c\n", *p);
return 0;} a
z
az
p c
'a'0x11320x1132
p g
'z'0x91A20x91A2
print “what p points to”
Writing Down Pointers
It is not only possible to read the values at the end of a pointer as with:
char c = 'a';char *p;
p = &c;printf("%c\n", *p);
char c = 'a';char *p;
p = &c;printf("%c\n", *p);
It is possible to write over the value at the end of a pointer:
char c = 'a';char *p;
p = &c;*p = 'b';printf("%c\n", *p);
char c = 'a';char *p;
p = &c;*p = 'b';printf("%c\n", *p);
p c
'a'0x11320x1132
'b'
make what p points to equal to ‘b’
Initialisation Warning!
The following code contains a horrible error:
#include <stdio.h>
int main(void){
short i = 13;short *p;
*p = 23;printf("%hi\n", *p);
return 0;}
#include <stdio.h>
int main(void){
short i = 13;short *p;
*p = 23;printf("%hi\n", *p);
return 0;}
p i
13?0x1212
Initialise Pointers!
Pointers are best initialised! A pointer may be declared and initialised in a
single step
short i = 13;short *p = &i;
short i = 13;short *p = &i;
This does NOT mean “make what p points to equal to the address of i”
It DOES mean “declare p as a pointer to a short int, make p equal to the address of i”
short *p = &i;short *p = &i; short *p = &i;short *p = &i;
short *p = &i;short *p = &i;
NULL
A special invalid pointer value exists #defined in various header files, called NULL
When assigned to a pointer, or when found in a pointer, it indicates the pointer is invalid
#include <stdio.h>
int main(void){
short i = 13;short *p = NULL;
if(p == NULL)printf("the pointer is invalid!\n");
elseprintf("the pointer points to %hi\n", *p);
return 0;}
#include <stdio.h>
int main(void){
short i = 13;short *p = NULL;
if(p == NULL)printf("the pointer is invalid!\n");
elseprintf("the pointer points to %hi\n", *p);
return 0;}
A World of Difference!
There is a great deal of difference between:
int i = 10, j = 14;int *p = &i;int *q = &j;
*p = *q;
int i = 10, j = 14;int *p = &i;int *q = &j;
*p = *q;
int i = 10, j = 14;int *p = &i;int *q = &j;
p = q;
int i = 10, j = 14;int *p = &i;int *q = &j;
p = q;
and:
p i
100x15A00x15A0
14
q j
0x15A40x15A4
14
p i
100x15A00x15A0
q j
0x15A40x15A4
14
0x15A4
Fill in the Gaps
int main(void){
int i = 10, j = 14, k;int *p = &i;int *q = &j;
*p += 1;
p = &k;
*p = *q;
p = q;
*p = *q;
return 0;}
int main(void){
int i = 10, j = 14, k;int *p = &i;int *q = &j;
*p += 1;
p = &k;
*p = *q;
p = q;
*p = *q;
return 0;}
i
0x2100
j
0x2104
k
0x1208
p
0x120B
q
0x1210
Type Mismatch
The compiler will not allow type mismatches when assigning to pointers, or to where pointers point
int i = 10, j = 14;int *p = &i;int *q = &j;
p = *q;*p = q;
int i = 10, j = 14;int *p = &i;int *q = &j;
p = *q;*p = q;
p i
100x15A00x15A0
q j
0x15A40x15A4
14
cannot write 0x15A4 into i cannot write
14 into p
Call by Value - Reminder
#include <stdio.h>
void change(int v);
int main(void){
int var = 5;
change(var);
printf("main: var = %i\n", var);
return 0;}
void change(int v){
v *= 100;printf("change: v = %i\n", v);
}
#include <stdio.h>
void change(int v);
int main(void){
int var = 5;
change(var);
printf("main: var = %i\n", var);
return 0;}
void change(int v){
v *= 100;printf("change: v = %i\n", v);
}change: v = 500main: var = 5
change: v = 500main: var = 5
the function was not able to alter “var”
the function is able to alter “v”
Call by Reference
#include <stdio.h>
void change(int* p);
int main(void){
int var = 5;
change(&var);
printf("main: var = %i\n", var);
return 0;}
void change(int* p){
*p *= 100;printf("change: *p = %i\n", *p);
}
#include <stdio.h>
void change(int* p);
int main(void){
int var = 5;
change(&var);
printf("main: var = %i\n", var);
return 0;}
void change(int* p){
*p *= 100;printf("change: *p = %i\n", *p);
}
change: *p = 500main: var = 500
change: *p = 500main: var = 500
change: p
main: var
5
0x1120
0x1120
0x1124
prototype “forces” us to pass a pointer
Pointers to Pointers
C allows pointers to any type It is possible to declare a pointer to a pointer
#include <stdio.h>
int main(void){
int i = 16;int *p = &i;int **pp;
pp = &p;printf("%i\n", **pp);
return 0;}
#include <stdio.h>
int main(void){
int i = 16;int *p = &i;int **pp;
pp = &p;printf("%i\n", **pp);
return 0;}
p
i16
0x2320
0x2320
0x2324
pp0x2324
0x2328
pp is a “pointer to” a “pointer to an int”
Review
int main(void){
int i = 10, j = 7, k;int *p = &i;int *q = &j;int *pp = &p;
**pp += 1;
*pp = &k;
**pp = *q;
i = *q***pp;
i = *q/**pp; /* headache? */;
return 0;}
int main(void){
int i = 10, j = 7, k;int *p = &i;int *q = &j;int *pp = &p;
**pp += 1;
*pp = &k;
**pp = *q;
i = *q***pp;
i = *q/**pp; /* headache? */;
return 0;}
p
i
pp
j k
q
Arrays in C
Declaring arrays Accessing elements Passing arrays into functions Using pointers to access arrays Strings The null terminator
Declaring Arrays
An array is a collection of data items (called elements) all of the same type
It is declared using a type, a variable name and a CONSTANT placed in square brackets
C always allocates the array in a single block of memory
The size of the array, once declared, is fixed forever - there is no equivalent of, for instance, the “redim” command in BASIC
Examples
#define SIZE 10int a[5]; /* a is an array of 5 ints */long int big[100]; /* big is 400 bytes! */double d[100]; /* but d is 800 bytes! */long double v[SIZE]; /* 10 long doubles, 100 bytes */
#define SIZE 10int a[5]; /* a is an array of 5 ints */long int big[100]; /* big is 400 bytes! */double d[100]; /* but d is 800 bytes! */long double v[SIZE]; /* 10 long doubles, 100 bytes */
int a[5] = { 10, 20, 30, 40, 50 };double d[100] = { 1.5, 2.7 };short primes[] = { 1, 2, 3, 5, 7, 11, 13 };long n[50] = { 0 };
int a[5] = { 10, 20, 30, 40, 50 };double d[100] = { 1.5, 2.7 };short primes[] = { 1, 2, 3, 5, 7, 11, 13 };long n[50] = { 0 };
int i = 7;const int c = 5;
int a[i];double d[c];short primes[];
int i = 7;const int c = 5;
int a[i];double d[c];short primes[];
all five elements initialised
first two elements initialised,
remaining ones set to zero
compiler fixes size at 7 elements
quickest way of settingALL elements to zero
Accessing Elements
The elements are accessed via an integer which ranges from 0..size-1
There is no bounds checking
int main(void){
int a[6];int i = 7;
a[0] = 59;a[5] = -10;a[i/2] = 2;
a[6] = 0;a[-1] = 5;
return 0;}
int main(void){
int a[6];int i = 7;
a[0] = 59;a[5] = -10;a[i/2] = 2;
a[6] = 0;a[-1] = 5;
return 0;}
0a
1
2
3
4
5
Array Names
There is a special and unusual property of array names in C
The name of an array is a pointer to the start of the array, i.e. the zeroth element, thus
a == &a[0]
int a[10];int *p;
float f[5]float *fp;
p = a; /* p = &a[0] */
fp = f; /* fp = &f[0] */
int a[10];int *p;
float f[5]float *fp;
p = a; /* p = &a[0] */
fp = f; /* fp = &f[0] */
ap
ffp
Passing Arrays to Functions
When an array is passed to a function a pointer to the zeroth element is passed across
The function may alter any element The corresponding parameter may be declared as
a pointer, or by using the following special syntax
int add_elements(int a[], int size){
int add_elements(int a[], int size){
int add_elements(int *p, int size){
int add_elements(int *p, int size){
Example
#include <stdio.h>
void sum(long [], int);
int main(void){
long primes[6] = { 1, 2,3, 5, 7, 11 };
sum(primes, 6);
printf("%li\n", primes[0]);
return 0;}
void sum(long a[], int sz){
int i;long total = 0;
for(i = 0; i < sz; i++)total += a[i];
a[0] = total;}
#include <stdio.h>
void sum(long [], int);
int main(void){
long primes[6] = { 1, 2,3, 5, 7, 11 };
sum(primes, 6);
printf("%li\n", primes[0]);
return 0;}
void sum(long a[], int sz){
int i;long total = 0;
for(i = 0; i < sz; i++)total += a[i];
a[0] = total;}
1
2
3
5
7
11
primes
a
sz 6
the total is written over element zero
provides bounds checking
Using Pointers
Pointers may be used to access array elements rather than using constructs involving “[ ]”
Pointers in C are automatically scaled by the size of the object pointed to when involved in arithmetic
long v[6] = { 1,2, 3,4,5,6 };
long *p;
p = v;printf("%ld\n", *p);p++;printf("%ld\n", *p);p += 4;printf("%ld\n", *p);
long v[6] = { 1,2, 3,4,5,6 };
long *p;
p = v;printf("%ld\n", *p);p++;printf("%ld\n", *p);p += 4;printf("%ld\n", *p);
v
p
p++p += 4
1000 1 2 3 4 5 6
126
126
10001004
10081012
10161020
Pointers Go Backwards Too
Scaling not only happens when addition is done, it happens with subtraction too
long v[6] = { 1,2, 3,4,5,6 };
long *p;
p = v + 5;printf("%ld\n", *p);p--;printf("%ld\n", *p);p -= 2;printf("%ld\n", *p);
long v[6] = { 1,2, 3,4,5,6 };
long *p;
p = v + 5;printf("%ld\n", *p);p--;printf("%ld\n", *p);p -= 2;printf("%ld\n", *p);
v
p
p--
1020 1 2 3 4 5 6
653
653
10001004
10081012
10161020
p-=2
Pointers May be Subtracted
When two pointers into the same array are subtracted C scales again, giving the number of array elements separating them
double d[7] = { 1.1, 2.2,3.3, 4.4, 5.5, 6.6, 7.7 };
double *p1;double *p2;
p1 = d + 1;p2 = d + 6;
printf("%i\n", p2 - p1);
double d[7] = { 1.1, 2.2,3.3, 4.4, 5.5, 6.6, 7.7 };
double *p1;double *p2;
p1 = d + 1;p2 = d + 6;
printf("%i\n", p2 - p1);
55
1.1 2.2 3.3 4.4 5.5 6.6 7.7
p1 p2
d
20002008
2016 2032 20482024 2040
20482008
Using Pointers - Example
1
2
3
5
7
11
primes
p
end 1024
1000
1004
#include <stdio.h>
long sum(long*, int);
int main(void){
long primes[6] = { 1, 2,3, 5, 7, 11 };
printf("%li\n", sum(primes, 6));
return 0;}
long sum(long *p, int sz){
long *end = p + sz;long total = 0;
while(p < end)total += *p++;
return total;}
#include <stdio.h>
long sum(long*, int);
int main(void){
long primes[6] = { 1, 2,3, 5, 7, 11 };
printf("%li\n", sum(primes, 6));
return 0;}
long sum(long *p, int sz){
long *end = p + sz;long total = 0;
while(p < end)total += *p++;
return total;}
1000
1008
1012
1016
1020
1024
* and ++
*p++ means:*p++ find the value at the end of the pointer*p++ increment the POINTER to point to the
next element(*p)++ means:
(*p)++ find the value at the end of the pointer(*p)++ increment the VALUE AT THE END OF THE POINTER (the pointer never moves)
*++p means:*++p increment the pointer*++p find the value at the end of the pointer
Which Notation?
An axiom of C states a[i] is equivalent to *(a + i)
ap
short a[8] = { 10, 20, 30, 40, 50, 60, 70, 80 };short *p = a;
printf("%i\n", a[3]);printf("%i\n", *(a + 3));printf("%i\n", *(p + 3));printf("%i\n", p[3]);printf("%i\n", 3[a]);
short a[8] = { 10, 20, 30, 40, 50, 60, 70, 80 };short *p = a;
printf("%i\n", a[3]);printf("%i\n", *(a + 3));printf("%i\n", *(p + 3));printf("%i\n", p[3]);printf("%i\n", 3[a]);
10 20 30 40 50 60 70 80
1000 1004 1008 10121002 1006 1010 1014
1000
4040404040
4040404040
Strings
C has no native string type, instead we use arrays of char
A special character, called a “null”, marks the end (don’t confuse this with the NULL pointer )
This may be written as ‘\0’ (zero not capital ‘o’) This is the only character whose ASCII value is
zero Depending on how arrays of characters are built,
we may need to add the null by hand, or the compiler may add it for us
Example
char first_name[5] = { 'J', 'o', 'h', 'n', '\0' };
char last_name[6] = "Minor";
char other[] = "Tony Blurt";
char characters[7] = "No null";
char first_name[5] = { 'J', 'o', 'h', 'n', '\0' };
char last_name[6] = "Minor";
char other[] = "Tony Blurt";
char characters[7] = "No null";
'J' 'o' 'h' 'n' 0first_name
'M' 'i' 'n' 'o' 'r'last_name 0
'T' 'o' 'n' 'y' 32other 'B' 'l' 'u' 'r' 't' 0
'N' 'o' 32 'n' 'u'characters 'l' 'l'
this special case specifically excludes the null terminator
Printing Strings
Strings may be printed by hand Alternatively printf supports “%s”
char other[] = "Tony Blurt";char other[] = "Tony Blurt";
char *p;
p = other;while(*p != '\0')
printf("%c", *p++);printf("\n");
char *p;
p = other;while(*p != '\0')
printf("%c", *p++);printf("\n");
int i = 0;
while(other[i] != '\0')printf("%c", other[i++]);
printf("\n");
int i = 0;
while(other[i] != '\0')printf("%c", other[i++]);
printf("\n");
printf("%s\n", other);printf("%s\n", other);
Null Really Does Mark the End!
#include <stdio.h>
int main(void){
char other[] = "Tony Blurt";
printf("%s\n", other);
other[4] = '\0';
printf("%s\n", other);
return 0;}
#include <stdio.h>
int main(void){
char other[] = "Tony Blurt";
printf("%s\n", other);
other[4] = '\0';
printf("%s\n", other);
return 0;} Tony Blurt
Tony
Tony BlurtTony
'T' 'o' 'n' 'y' 32other 'B' 'l' 'u' 'r' 't' 0
even though the rest of the data is still there, printf will NOT move past the null terminator
Assigning to Strings
Strings may be initialised with “=”, but not assigned to with “=”
Remember the name of an array is a CONSTANT pointer to the zeroth element
#include <stdio.h>#include <string.h>
int main(void){
char who[] = "Tony Blurt";
who = "John Minor";
strcpy(who, "John Minor");
return 0;}
#include <stdio.h>#include <string.h>
int main(void){
char who[] = "Tony Blurt";
who = "John Minor";
strcpy(who, "John Minor");
return 0;}
Pointing to Strings
To save us declaring many character arrays to store strings, the compiler can store them directly in the data segment
We need only declare a pointer The compiler may recycle some of these strings,
therefore we must NOT alter any of the characters
char *p = "Data segment!!";char *q = "nt!!";
char *p = "Data segment!!";char *q = "nt!!";
'D' 'a' 't' 'a' 32 's' 'e' 'g' 'm' 'e' 'n' 't' '!' '!' 0
p 0xF100
0xF100 0xF10A
q 0xF10A
Example
#include <stdio.h>
int main(void){
char *p = "a string in the data segment\n";
"a second string in the data segment\n";
printf("a third string in the data segment\n");
printf("%s", p);
printf(p);
return 0;}
#include <stdio.h>
int main(void){
char *p = "a string in the data segment\n";
"a second string in the data segment\n";
printf("a third string in the data segment\n");
printf("%s", p);
printf(p);
return 0;}
this utterly pointless statement causes the compiler to store the characters, unfortunately we forget to save the address
a third string in the data segmenta string in the data segmenta string in the data segment
a third string in the data segmenta string in the data segmenta string in the data segment
Multidimensional Arrays
C does not support multidimensional arrays However, C does support arrays of any type
including arrays of arrays
float rainfall[12][365];float rainfall[12][365]; “rainfall” is an array of 12 arrays of 365 float
short exam_marks[500][10];short exam_marks[500][10]; “exam_marks” is an array of 500 arrays of 10 short int
const int brighton = 7;int day_of_year = 238;
rainfall[brighton][day_of_year] = 0.0F;
const int brighton = 7;int day_of_year = 238;
rainfall[brighton][day_of_year] = 0.0F;
Review
#include <stdio.h>
int main(void){
int i;int a[10];
for(i = 0; i <= 10; i++) {printf("%d\n", i);a[i] = 0;
}
return 0;}
#include <stdio.h>
int main(void){
int i;int a[10];
for(i = 0; i <= 10; i++) {printf("%d\n", i);a[i] = 0;
}
return 0;}
How many times does the following program loop?
Summary
Arrays are declared with a type, a name, “[ ]” and a CONSTANT
Access to elements by array name, “[ ]” and an integer
Arrays passed into functions by pointer Pointer arithmetic Strings - arrays of characters with a null
terminator Sometimes compiler stores null for us (when
double quotes are used) otherwise we have to store it ourselves
Structures in C
Concepts Creating a structure template Using the template to create an instance Initializing an instance Accessing an instance’s members Passing instances to functions Linked lists
Concepts
A structure is a collection of one of more variables grouped together under a single name for convenient handling
The variables in a structure are called members and may have any type, including arrays or other structures
The steps are:– set-up a template (blueprint) to tell the compiler how to build
the structure
– Use the template to create as many instances of the structure as desired
– Access the members of an instance as desired
Setting up the Template
Structure templates are created by using the struct keyword
struct Book{
char title[80];char author[80];float price;char isbn[20];
};
struct Book{
char title[80];char author[80];float price;char isbn[20];
};
struct Date{
int day;int month;int year;
};
struct Date{
int day;int month;int year;
};
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
struct Library_book{
struct Book b;struct Date due;struct Library_member *who;
};
struct Library_book{
struct Book b;struct Date due;struct Library_member *who;
};
Creating Instances
Having created the template, an instance (or instances) of the structure may be declared
struct Date{
int day;int month;int year;
} today, tomorrow;
struct Date next_monday;
struct Date next_week[7];
struct Date{
int day;int month;int year;
} today, tomorrow;
struct Date next_monday;
struct Date next_week[7];
an array of 7 date instances
instances must be declared before the ‘;’ ...
... or “struct Date” has to be repeated
Initialising Instances
Structure instances may be initialised using braces (as with arrays)
int primes[7] = { 1, 2, 3, 5, 7, 11, 13 };
struct Date bug_day = { 1, 1, 2000 };
struct Book k_and_r = {"The C Programming Language 2nd edition","Brian W. Kernighan and Dennis M. Ritchie",31.95,"0-13-110362-8"
};
int primes[7] = { 1, 2, 3, 5, 7, 11, 13 };
struct Date bug_day = { 1, 1, 2000 };
struct Book k_and_r = {"The C Programming Language 2nd edition","Brian W. Kernighan and Dennis M. Ritchie",31.95,"0-13-110362-8"
}; struct Book{
char title[80];char author[80];float price;char isbn[20];
};
struct Book{
char title[80];char author[80];float price;char isbn[20];
};
Structures Within Structures
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
struct Library_member m = {"Arthur Dent","16 New Bypass",42,{ 0.10, 2.58, 0.13, 1.10 },{ 18, 9, 1959 },{ 1, 4, 1978 }
};
struct Library_member m = {"Arthur Dent","16 New Bypass",42,{ 0.10, 2.58, 0.13, 1.10 },{ 18, 9, 1959 },{ 1, 4, 1978 }
};
initialises first 4 elements of array “fines”, remainder are initialised to 0.0
initialises day, month and year of “dob”
initialises day, month and year of “enrolled”
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
struct Library_member{
char name[80];char address[200];long member_number;float fines[10];struct Date dob;struct Date enrolled;
};
Accessing Members
Members are accessed using the instance name, “.” and the member name
printf("name = %s\n", m.name);printf("membership number = %li\n", m.member_number);
printf("fines: ");for(i = 0; i < 10 && m.fines[i] > 0.0; i++)
printf("£%.2f ", m.fines[i]);
printf("\njoined %i/%i/%i\n", m.enrolled.day,m.enrolled.month, m.enrolled.year);
printf("name = %s\n", m.name);printf("membership number = %li\n", m.member_number);
printf("fines: ");for(i = 0; i < 10 && m.fines[i] > 0.0; i++)
printf("£%.2f ", m.fines[i]);
printf("\njoined %i/%i/%i\n", m.enrolled.day,m.enrolled.month, m.enrolled.year);
struct Library_member m;struct Library_member m;
Unusual Properties
Structures have some very “un-C-like” properties, certainly when considering how arrays are handled
Arrays Structures
Name is pointer to the structure itselfzeroth element
Passed to functions by pointer value or pointer
Returned from functions no way by value or pointer
May be assigned with “=” no way yes
Instances may be Assigned
Two structure instances may be assigned to one another via “=”
All the members of the instance are copied (including arrays or other structures)
struct Library_member m = {"Arthur Dent",.....
};struct Library_member tmp;
tmp = m;
struct Library_member m = {"Arthur Dent",.....
};struct Library_member tmp;
tmp = m;
copies array “name”, array “address”, long integer “member_number”, array “fines”, Date structure “dob” and Date structure “enrolled”
Passing Instances to Functions
An instance of a structure may be passed to a function by value or by pointer
Pass by value becomes less and less efficient as the structure size increases
Pass by pointer remains efficient regardless of the structure size
void by_value(struct Library_member);void by_reference(struct Library_member *);
by_value(m);by_reference(&m);
void by_value(struct Library_member);void by_reference(struct Library_member *);
by_value(m);by_reference(&m);
compiler writes a pointer (4 bytes?) onto the stack
compiler writes 300+ bytes onto the stack
Pointers to Structures
Passing pointers to structure instances is more efficient
Dealing with an instance at the end of a pointer is not so straightforward!
void member_display(struct Library_member *p){
printf("name = %s\n", (*p).name);printf("membership number = %li\n", (*p).member_number);
printf("fines: ");for(i = 0; i < 10 && (*p).fines[i] > 0.0; i++)
printf("£%.2f ", (*p).fines[i]);
printf("\njoined %i/%i/%i\n", (*p).enrolled.day,(*p).enrolled.month, (*p).enrolled.year);
}
void member_display(struct Library_member *p){
printf("name = %s\n", (*p).name);printf("membership number = %li\n", (*p).member_number);
printf("fines: ");for(i = 0; i < 10 && (*p).fines[i] > 0.0; i++)
printf("£%.2f ", (*p).fines[i]);
printf("\njoined %i/%i/%i\n", (*p).enrolled.day,(*p).enrolled.month, (*p).enrolled.year);
}
Why (*p).name ?
The messy syntax is needed because “.” has higher precedence than “*”, thus:
*p.name
means “what p.name points to” (a problem because there is no structure instance “p”)
As Kernighan and Ritchie foresaw pointers and structures being used frequently they invented a new operator
p->name = (*p).name
Using p->name
Now dealing with the instance at the end of the pointer is more straightforward
void member_display(struct Library_member *p){
printf("name = %s\n", p->name);printf("address = %s\n", p->address);printf("membership number = %li\n", p->member_number);
printf("fines: ");for(i = 0; i < 10 && p->fines[i] > 0.0; i++)
printf("£%.2f ", p->fines[i]);
printf("\njoined %i/%i/%i\n", p->enrolled.day,p->enrolled.month, p->enrolled.year);
}
void member_display(struct Library_member *p){
printf("name = %s\n", p->name);printf("address = %s\n", p->address);printf("membership number = %li\n", p->member_number);
printf("fines: ");for(i = 0; i < 10 && p->fines[i] > 0.0; i++)
printf("£%.2f ", p->fines[i]);
printf("\njoined %i/%i/%i\n", p->enrolled.day,p->enrolled.month, p->enrolled.year);
}
Pass by Reference - Warning
Although pass by reference is more efficient, the function can alter the structure (perhaps inadvertently)
Use a pointer to a constant structure instead
void member_display(struct Library_member *p){
printf("fines: ");for(i = 0; i < 10 && p->fines[i] = 0.0; i++)
printf("£%.2f ", p->fines[i]);}
void member_display(struct Library_member *p){
printf("fines: ");for(i = 0; i < 10 && p->fines[i] = 0.0; i++)
printf("£%.2f ", p->fines[i]);}
void member_display(const struct Library_member *p){
....}
void member_display(const struct Library_member *p){
....}
function alters the library member instance
Returning Structure Instances
Structure instances may be returned by value from functions
This can be as inefficient as with pass by value Sometimes it is convenient!
struct Complex add(struct Complex a, struct Complex b){
struct Complex result = a;
result.real_part += b.real_part;result.imag_part += b.imag_part;
return result;}
struct Complex add(struct Complex a, struct Complex b){
struct Complex result = a;
result.real_part += b.real_part;result.imag_part += b.imag_part;
return result;} struct Complex c1 = { 1.0, 1.1 };
struct Complex c2 = { 2.0, 2.1 };struct Complex c3;
c3 = add(c1, c2); /* c3 = c1 + c2 */
struct Complex c1 = { 1.0, 1.1 };struct Complex c2 = { 2.0, 2.1 };struct Complex c3;
c3 = add(c1, c2); /* c3 = c1 + c2 */
Linked Lists
A linked list node containing a single forward pointer may be declared as follows
struct Node {int data; /* or whatever */struct Node *next_in_line;
};
struct Node {int data; /* or whatever */struct Node *next_in_line;
};
pointer to next Node structure
A linked list node containing a forward and a backward pointer may be declared as follows
struct Node {int data;struct Node *next_in_line;struct Node *previous_in_line;
};
struct Node {int data;struct Node *next_in_line;struct Node *previous_in_line;
};pointer to previous Node structure
pointer to next Node structure
Example
#include <stdio.h>
struct Node {char name[10];struct Node *next_in_line;
};
struct Node a1 = { "John", NULL };struct Node a2 = { "Harriet", &a1 },struct Node a3 = { "Claire", &a2 }struct Node a4 = { "Tony", &a3 };
#include <stdio.h>
struct Node {char name[10];struct Node *next_in_line;
};
struct Node a1 = { "John", NULL };struct Node a2 = { "Harriet", &a1 },struct Node a3 = { "Claire", &a2 }struct Node a4 = { "Tony", &a3 };
Tony\0
a4
0x1012
Claire\0
a3
0x1020
Harriet\0
a2
0x102E
John\0
a1
0x1032
0x1020 0x102E 0x1032 NULL
Printing the List
The list may be printed with the following code:
struct Node * current = &a4;
while(current != NULL) {printf("%s\n", current->name);current = current->next_in_line;
}
struct Node * current = &a4;
while(current != NULL) {printf("%s\n", current->name);current = current->next_in_line;
}
current
0x1012
Tony\0
a4
0x1012
Claire\0
0x1020
Harriet\0
0x102E
John\0
0x1032
0x1020 0x102E 0x1032 NULL
Summary
Creating structure templates using struct Creating and initialising instances Accessing members Passing instances to functions by value and by
reference A new operator: “->” Return by value Linked lists
Introduction
Up until now we have seen straightforward declarations:
void member_display(const struct Library_member *p);void member_display(const struct Library_member *p);
However, they can become much worse:
int *p[15];float (*pfa)[23];long (*f)(char, int);double *(*(*n)(void))[5];
int *p[15];float (*pfa)[23];long (*f)(char, int);double *(*(*n)(void))[5];
long sum;int* p;
long sum;int* p;
Plus a few trickier ones:
SOAC
Find the variable being declared Spiral Outwards Anti Clockwise On meeting: say:
* pointer to
[ ] array of
( ) function taking .... and returning
Remember to read “struct S”, “union U” or “enum E” all at once
Remember to read adjacent collections of [ ] [ ] all at once
Example 3.
What is “short **ab[5][10]” ?
short * * ab [5][10] ;
ab is an array of 5 arrays of 10 arrays of pointers to pointers to short int
Example 4.
What is “long * f(int, float)” ?
long * f (int, float) ;
f is a function taking an int and a float returning a pointer to a long int
Example 5.
What is “int (*pf)(void)” ?
int ( * pf ) (void) ;
pf is a pointer to a function taking no parameters and returning an int
Example 6.
What is “struct Book (*fpa[8])(void)” ?
struct Book ( * fpa[8] ) (void) ;
fpa is an array of 8 pointers to functions, taking no parameters, returning Book structures
Example 7.
What is “char (*(*fprp)(void))[6]” ?
char ( * ( * fprp ) (void) ) [6] ;
fprp is a pointer to a function taking no parameters returning a pointer to an array of 6 char
Example 8.
What is “int * (*(*ptf)(int))(char)” ?
int * ( * ( * ptf ) (int) ) (char) ;
ptf is a pointer to a function, taking an integer, returning a pointer to a function, taking a char, returning a pointer to an int
typedef
It doesn’t have to be this difficult! The declaration can be broken into simpler steps
by using typedef To tackle typedef, pretend it isn’t there and read
the declaration as for a variable When finished remember that a type has been
declared, not a variable
Example 1 Revisited
Simplify “int * p[15]”
typedef int * pti ;
pti p[15];
pti is a pointer to an int
p is an array of 15 pointer to int
Example 3 Revisited
Simplify “short **ab[5][10]”
typedef short * * pt_pt_s ;
ab is an array of 10 arrays of 5 pointers to pointers to short
typedef pt_pt_s ao5[5];
ao5 ab[10];
ao5 is an array of 5 pointers to pointers to short
pt_pt_s is a pointer to a pointer to a short
Example 5 Revisited
Simplify “int (*pf)(void)”
typedef int fri(void); fri * pf ;
fri is a function, taking no parameters, returning an int
pf is a pointer to a function, taking no parameters,
returning an int
Example 6 Revisited
Simplify “struct Book (*fpa[8])(void)”
typedef struct Book f(void);
f is a function, taking no parameters, returning a
Book structure
typedef f * fp ;
fp is a pointer to a function, taking no parameters,
returning a Book structure
fp fpa[8];
fpa is an array of 8 pointers to functions, taking no parameters, returning a
Book structure
Example 7 Revisited
Simplify “char (*(*fprp)(void))[6]”
typedef char ( * pta6c ) [6] ;
pta6c is a pointer to an array of 6 char
typedef pta6c f(void);
f is a function, taking no parameters, returning a pointer to an array of 6
char
f * fprp ;
fprp is a pointer to a function, taking no
parameters, returning a pointer to an array of 6
char
Example 8 Revisited
Simplify “int * (*(*ptf)(int))(char)”
typedef int * pti ;
pti is a pointer to an int
typedef pti f(char);
f is a function, taking a char, returning a pointer
to an int
typedef f * ptfri ;
ptfri is a pointer to a function, taking a char,
returning a pointer to an int
ptfri ( * ptf )(int) ;
ptf is a pointer to a function, taking int, returning a pointer to a function, taking a
char, returning a pointer to an int
Handling Files in C
Streams stdin, stdout, stderr Opening files When things go wrong - perror Copying files Accessing the command line Dealing with binary files
Introduction
File handling is not built into the C language itself It is provided by The Standard Library (via a set
of routines invariably beginning with “f”) Covered by The Standard, the routines will
always be there and work the same way, regardless of hardware/operating system
Files are presented as a sequence of characters It is easy to move forwards reading/writing
characters, it is less easy (though far from impossible) to go backwards
Streams
Before a file can be read or written, a data structure known as a stream must be associated with it
A stream is usually a pointer to a structure (although it isn’t necessary to know this)
There are three streams opened by every C program, stdin, stdout and stderr
stdin (standard input) is connected to the keyboard and may be read from
stdout (standard output) and stderr (standard error) are connected to the screen and may be written to
What is a Stream?
Although implementations vary, a stream creates a buffer between the program running in memory and the file on the disk
This reduces the program’s need to access slow hardware devices
Characters are silently read a block at a time into the buffer, or written a block at a time to the file
a b c d ef gh i j k l
output stream
a b c de f g hi j
input stream
Why stdout and stderr?
There are two output streams because of redirection, supported by Unix, DOS, OS/2 etc.
#include <stdio.h>
int main(void){
printf("written to stdout\n");fprintf(stderr, "written to stderr\n");
return 0;}
#include <stdio.h>
int main(void){
printf("written to stdout\n");fprintf(stderr, "written to stderr\n");
return 0;}
C:> outprogwritten to stderrwritten to stdoutC:> outprog > file.txtwritten to stderrC:> type file.txtwritten to stdout
C:> outprogwritten to stderrwritten to stdoutC:> outprog > file.txtwritten to stderrC:> type file.txtwritten to stdout
output written to stderr first because it is unbuffered
stdin is Line Buffered
Characters typed at the keyboard are buffered until Enter/Return is pressed
#include <stdio.h>
int main(void){
int ch;
while((ch = getchar()) != EOF)printf("read '%c'\n", ch);
printf("EOF\n");
return 0;}
#include <stdio.h>
int main(void){
int ch;
while((ch = getchar()) != EOF)printf("read '%c'\n", ch);
printf("EOF\n");
return 0;}
C:> inprogabcread 'a'read 'b'read 'c'read ''dread 'd'read ''^ZEOFC:>
C:> inprogabcread 'a'read 'b'read 'c'read ''dread 'd'read ''^ZEOFC:>declared as an int, even though
we are dealing with characters
Opening Files
Files are opened and streams created with the fopen function
FILE* fopen(const char* name, const char* mode);FILE* fopen(const char* name, const char* mode);
#include <stdio.h>
int main(void){
FILE* in;FILE* out;FILE* append;
in = fopen("autoexec.bat", "r");out = fopen("autoexec.bak", "w");append = fopen("config.sys", "a");
#include <stdio.h>
int main(void){
FILE* in;FILE* out;FILE* append;
in = fopen("autoexec.bat", "r");out = fopen("autoexec.bak", "w");append = fopen("config.sys", "a");
streams, you’ll need one for each
file you want open
#include <stdio.h>
int main(void){
FILE* in;
if((in = fopen("autoexec.bat", "r")) == NULL) {fprintf(stderr, "open of autoexec.bat failed ");perror("because");return 1;
}
#include <stdio.h>
int main(void){
FILE* in;
if((in = fopen("autoexec.bat", "r")) == NULL) {fprintf(stderr, "open of autoexec.bat failed ");perror("because");return 1;
}
Dealing with Errors
fopen may fail for one of many reasons, how to tell which?
void perror(const char* message);void perror(const char* message);
open of autoexec.bat failed because: No such file or directoryopen of autoexec.bat failed because: No such file or directory
File Access Problem
Can you see why the following will ALWAYS fail, despite the file existing and being fully accessible?
if((in = fopen("C:\autoexec.bat", "r")) == NULL) {fprintf(stderr, "open of autoexec.bat failed ");perror("because");return 1;
}
if((in = fopen("C:\autoexec.bat", "r")) == NULL) {fprintf(stderr, "open of autoexec.bat failed ");perror("because");return 1;
}
C:> dir C:\autoexec.bat Volume in drive C is MS-DOS_62 Directory of C:\
autoexec bat 805 29/07/90 8:15 1 file(s) 805 bytes 1,264,183,808 bytes freeC:> myprogopen of autoexec.bat failed because: No such file or directory
C:> dir C:\autoexec.bat Volume in drive C is MS-DOS_62 Directory of C:\
autoexec bat 805 29/07/90 8:15 1 file(s) 805 bytes 1,264,183,808 bytes freeC:> myprogopen of autoexec.bat failed because: No such file or directory
Displaying a File
#include <stdio.h>
int main(void){
char in_name[80];FILE *in_stream;int ch;
printf("Display file: ");scanf("%79s", in_name);
if((in_stream = fopen(in_name, "r")) == NULL) {fprintf(stderr, "open of %s for reading failed ", in_name);perror("because");return 1;
}
while((ch = fgetc(in_stream)) != EOF)putchar(ch);
fclose(in_stream);
return 0;}
#include <stdio.h>
int main(void){
char in_name[80];FILE *in_stream;int ch;
printf("Display file: ");scanf("%79s", in_name);
if((in_stream = fopen(in_name, "r")) == NULL) {fprintf(stderr, "open of %s for reading failed ", in_name);perror("because");return 1;
}
while((ch = fgetc(in_stream)) != EOF)putchar(ch);
fclose(in_stream);
return 0;}
Example - Copying Files#include <stdio.h>
int main(void){
char in_name[80], out_name[80];FILE *in_stream, *out_stream;int ch;
printf("Source file: "); scanf("%79s", in_name);if((in_stream = fopen(in_name, "r")) == NULL) {
fprintf(stderr, "open of %s for reading failed ", in_name);perror("because");return 1;
}
printf("Destination file: "); scanf("%79s", out_name);if((out_stream = fopen(out_name, "w")) == NULL) {
fprintf(stderr, "open of %s for writing failed ", out_name);perror("because");return 1;
}
while((ch = fgetc(in_stream)) != EOF)fputc(ch, out_stream);
fclose(in_stream);fclose(out_stream);
return 0;}
#include <stdio.h>
int main(void){
char in_name[80], out_name[80];FILE *in_stream, *out_stream;int ch;
printf("Source file: "); scanf("%79s", in_name);if((in_stream = fopen(in_name, "r")) == NULL) {
fprintf(stderr, "open of %s for reading failed ", in_name);perror("because");return 1;
}
printf("Destination file: "); scanf("%79s", out_name);if((out_stream = fopen(out_name, "w")) == NULL) {
fprintf(stderr, "open of %s for writing failed ", out_name);perror("because");return 1;
}
while((ch = fgetc(in_stream)) != EOF)fputc(ch, out_stream);
fclose(in_stream);fclose(out_stream);
return 0;}
Convenience Problem
Although our copy file program works, it is not as convenient as the “real thing”
C:> copyprogSource file: \autoexec.batDestination file: \autoexec.bakC:> dir C:\autoexec.* Volume in drive C is MS-DOS_62 Directory of C:\
autoexec bak 805 31/12/99 12:34autoexec bat 805 29/07/90 8:15 2 file(s) 1610 bytes 1,264,183,003 bytes freeC:> copyprog \autoexec.bat \autoexec.000Source file:
C:> copyprogSource file: \autoexec.batDestination file: \autoexec.bakC:> dir C:\autoexec.* Volume in drive C is MS-DOS_62 Directory of C:\
autoexec bak 805 31/12/99 12:34autoexec bat 805 29/07/90 8:15 2 file(s) 1610 bytes 1,264,183,003 bytes freeC:> copyprog \autoexec.bat \autoexec.000Source file:
program still prompts despite begin given file names on the command line
Accessing the Command Line
The command line may be accessed via two parameters to main, by convention these are called “argc” and “argv”
The first is a count of the number of words - including the program name itself
The second is an array of pointers to the words
int main(int argc, char *argv[])int main(int argc, char *argv[])
argc argv3
NULL
c o p y p r o g . e x e \0
\ a u t o e x e c . b a t \0
\ a u t o e x e c . 0 0 0 \0
Example
#include <stdio.h>
int main(int argc, char *argv[]){
int j;
for(j = 0; j < argc; j++)printf("argv[%i] = \"%s\"\n", j, argv[j]);
return 0;}
#include <stdio.h>
int main(int argc, char *argv[]){
int j;
for(j = 0; j < argc; j++)printf("argv[%i] = \"%s\"\n", j, argv[j]);
return 0;}
C:> argprog one two threeargv[0] = "C:\cct\course\cprog\files\slideprog\argprog.exe"argv[1] = "one"argv[2] = "two"argv[3] = "three"
C:> argprog one two threeargv[0] = "C:\cct\course\cprog\files\slideprog\argprog.exe"argv[1] = "one"argv[2] = "two"argv[3] = "three"
Useful Routines
File reading routines:
int fscanf(FILE* stream, const char* format, ...);int fgetc(FILE* stream);char* fgets(char* buffer, int size, FILE* stream);
int fscanf(FILE* stream, const char* format, ...);int fgetc(FILE* stream);char* fgets(char* buffer, int size, FILE* stream);
File writing routines:
int fprintf(FILE* stream, const char* format, ...);int fputc(int ch, FILE* stream);int fputs(const char* buffer, FILE* stream);
int fprintf(FILE* stream, const char* format, ...);int fputc(int ch, FILE* stream);int fputs(const char* buffer, FILE* stream);
Example
long l1, l2;int j, ch;double d;float f;char buf[200];
in = fopen("in.txt", "r") .... out = fopen("out.txt", "w") ....
fscanf(in, "%lf|%li:%li/%i", &d, &l1, &l2, &j);fprintf(out, "%li:%i:%.2lf\n", l1, j, d);
fgetc(in);
fgets(buf, sizeof(buf), in);fputs(buf, out);
long l1, l2;int j, ch;double d;float f;char buf[200];
in = fopen("in.txt", "r") .... out = fopen("out.txt", "w") ....
fscanf(in, "%lf|%li:%li/%i", &d, &l1, &l2, &j);fprintf(out, "%li:%i:%.2lf\n", l1, j, d);
fgetc(in);
fgets(buf, sizeof(buf), in);fputs(buf, out);
example input
28.325|9000000:68000/1328.325|9000000:68000/13
write that line to the output file (null terminator provided by fgets tells fputs how long the line was)
read next line, or next 199 characters,
whichever is less
ignore next character in input file (newline?)
9000000:13:28.339000000:13:28.33
Binary Files
The Standard Library also allows binary files to be manipulated– “b” must be added into the fopen options
– Character translation is disabled
– Random access becomes easier
– Finding the end of file can become more difficult
– Data is read and written in blocks
size_t fread(void* p, size_t size, size_t n, FILE* stream);size_t fwrite(const void* p, size_t size, size_t n, FILE* stream);
int fseek(FILE* stream, long offset, int whence);long ftell(FILE* stream);void rewind(FILE* stream);
int fgetpos(FILE* stream, fpos_t* pos);int fsetpos(FILE* stream, const fpos_t* pos);
size_t fread(void* p, size_t size, size_t n, FILE* stream);size_t fwrite(const void* p, size_t size, size_t n, FILE* stream);
int fseek(FILE* stream, long offset, int whence);long ftell(FILE* stream);void rewind(FILE* stream);
int fgetpos(FILE* stream, fpos_t* pos);int fsetpos(FILE* stream, const fpos_t* pos);
Example
double d;long double lda[35];fpos_t where;
in = fopen("binary.dat", "rb");out = fopen("binnew.dat", "wb");
fread(&d, sizeof(d), 1, in);
fgetpos(in, &where);fread(lda, sizeof(lda), 1, in);
fsetpos(in, &where);fread(lda, sizeof(long double), 35, in);
fwrite(lda, sizeof(long double), 20, out);
fseek(in, 0L, SEEK_END);
double d;long double lda[35];fpos_t where;
in = fopen("binary.dat", "rb");out = fopen("binnew.dat", "wb");
fread(&d, sizeof(d), 1, in);
fgetpos(in, &where);fread(lda, sizeof(lda), 1, in);
fsetpos(in, &where);fread(lda, sizeof(long double), 35, in);
fwrite(lda, sizeof(long double), 20, out);
fseek(in, 0L, SEEK_END);
read one chunk of 8 bytes
read one chunk of 350 bytes
read 35 chunks of 10 bytes
remember current position in file
return to previous position
write 20 long doubles from ldamove to end of binary.dat
Summary
Streams stdin, stdout, stderr fopen opening text files functions: perror, fprintf, fscanf, fgetc, fputc
variables: argc, argv “b” option to fopen to open binary files functions: fread, fwrite, fseek, ftell
Unions
A union is a variable which, at different times, may hold objects of different types and sizes
struct S{
short s;long l;double d;char c;
} s;
s.s = 10;s.l = 10L;s.d = 10.01;s.c = '1';
struct S{
short s;long l;double d;char c;
} s;
s.s = 10;s.l = 10L;s.d = 10.01;s.c = '1';
union U{
short s;long l;double d;char c;
} u;
u.s = 10;u.l = 10L;u.d = 10.01;u.c = '1';
union U{
short s;long l;double d;char c;
} u;
u.s = 10;u.l = 10L;u.d = 10.01;u.c = '1';
s
u
Remembering
It is up to the programmer to remember what type a union currently holds
Unions are most often used in structures where a member records the type currently stored
struct preprocessor_const{
char* name;int stored;union{
long lval;double dval;char* sval;
} u;};
struct preprocessor_const{
char* name;int stored;union{
long lval;double dval;char* sval;
} u;};
#define N_SIZE 10#define PI 3.1416
#define N_SIZE 10#define PI 3.1416
struct preprocessor_const s[10000];
s[0].name = "N_SIZE";s[0].u.lval = 10L;s[0].stored = STORED_LONG;
s[1].name = "PI"; s[1].u.dval = 3.1416;s[1].stored = STORED_DOUBLE;
struct preprocessor_const s[10000];
s[0].name = "N_SIZE";s[0].u.lval = 10L;s[0].stored = STORED_LONG;
s[1].name = "PI"; s[1].u.dval = 3.1416;s[1].stored = STORED_DOUBLE;
Enumerated Types
Enumerated types provide an automated mechanism for generating named constants
enum day { sun, mon, tue,wed, thu, fri, sat };
enum day today = sun;
if(today == mon)....
enum day { sun, mon, tue,wed, thu, fri, sat };
enum day today = sun;
if(today == mon)....
#define sun 0#define mon 1#define tue 2#define wed 3#define thu 4#define fri 5#define sat 6
int today = sun;
if(today == mon)....
#define sun 0#define mon 1#define tue 2#define wed 3#define thu 4#define fri 5#define sat 6
int today = sun;
if(today == mon)....
Using Different Constants
The constants used may be specified
enum day { sun = 5, mon, tue, wed, thu, fri, sat };
enum direction { north = 0, east = 90, south = 180,west = 270 };
enum day { sun = 5, mon, tue, wed, thu, fri, sat };
enum direction { north = 0, east = 90, south = 180,west = 270 };
What you see is all you get! There are no successor or predecessor functions
The Preprocessor
Preprocessor commands start with ‘#’ which may optionally be surrounded by spaces and tabs
The preprocessor allows us to:
– include files
– define, test and compare constants
– write macros
– debug
Including Files
The #include directive causes the preprocessor to “edit in” the entire contents of another file
#define JAN 1#define FEB 2#define MAR 3
#define PI 3.1416
double my_global;
#define JAN 1#define FEB 2#define MAR 3
#define PI 3.1416
double my_global;
mydefs.h
#include "mydefs.h"
double angle = 2 * PI;printf("%s", month[FEB]);
#include "mydefs.h"
double angle = 2 * PI;printf("%s", month[FEB]);
myprog.c
#define JAN 1#define FEB 2#define MAR 3
#define PI 3.1416
double my_global;
double angle = 2 * 3.1416;printf("%s", month[2]);
#define JAN 1#define FEB 2#define MAR 3
#define PI 3.1416
double my_global;
double angle = 2 * 3.1416;printf("%s", month[2]);
myprog.i
Pathnames
Full pathnames may be used, although this is not recommended
#include "C:\cct\course\cprog\misc\slideprog\header.h"#include "C:\cct\course\cprog\misc\slideprog\header.h"
The “I” directive to your local compiler allows code to be moved around much more easily
#include "header.h"#include "header.h"
cc -I c:\cct\course\cprog\misc\slideprog myprog.ccc -I c:\cct\course\cprog\misc\slideprog myprog.c
Preprocessor Constants
Constants may be created, tested and removed
#if !defined(SUN)#define SUN 0#endif
#if SUN == MON#undef SUN#endif
#if !defined(SUN)#define SUN 0#endif
#if SUN == MON#undef SUN#endif
if “SUN” is not defined, then begin
define “SUN” as zero
end
#if SUN > SAT && SUN > MON#if SUN > SAT && SUN > MON
#if WED > 0 || SUN < 3#if WED > 0 || SUN < 3
#if TUE#if TUE if “TUE” is defined with a non zero value
if “WED” is greater than zero or “SUN” is less than 3
if “SUN” is greater than “SAT” and “SUN” is greater than “MON”
if “SUN” and “MON” are equal, then begin
remove definition of “SUN”
end
Avoid Temptation!
The following attempt to write Pascal at the C compiler will ultimately lead to tears
#define begin {#define end ;}#define if if(#define then )#define integer int
integer i;
if i > 0 then begini = 17
end
#define begin {#define end ;}#define if if(#define then )#define integer int
integer i;
if i > 0 then begini = 17
end
int i;
if( i > 0 ) {i = 17
;}
int i;
if( i > 0 ) {i = 17
;}
Preprocessor Macros
The preprocessor supports a macro facility which should be used with care
#define MAX(A,B) A > B ? A : B#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
int i = 10, j = 12, k;
k = MAX(i, j); printf("k = %i\n", k);k = MAX(j, i) * 2; printf("k = %i\n", k);k = MIN(i, j) * 3; printf("k = %i\n", k);k = MIN(i--, j++); printf("i = %i\n", i);
#define MAX(A,B) A > B ? A : B#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
int i = 10, j = 12, k;
k = MAX(i, j); printf("k = %i\n", k);k = MAX(j, i) * 2; printf("k = %i\n", k);k = MIN(i, j) * 3; printf("k = %i\n", k);k = MIN(i--, j++); printf("i = %i\n", i); k = 12
k = 12k = 30i = 8
k = 12k = 12k = 30i = 8
A Debugging Aid
Several extra features make the preprocessor an indespensible debugging tool
#define GOT_HERE printf("reached %i in %s\n", \_ _LINE_ _, _ _FILE_ _)
#define SHOW(E, FMT) printf(#E " = " FMT "\n", E)
#define GOT_HERE printf("reached %i in %s\n", \_ _LINE_ _, _ _FILE_ _)
#define SHOW(E, FMT) printf(#E " = " FMT "\n", E)
printf("i = %x\n", i);printf("i = %x\n", i);
printf("reached %i in %s\n", 17, "mysource.c");printf("reached %i in %s\n", 17, "mysource.c");
GOT_HERE;SHOW(i, "%x");SHOW(f/29.5, "%lf");
GOT_HERE;SHOW(i, "%x");SHOW(f/29.5, "%lf");
printf("f/29.5 = %lf\n", f/29.5);printf("f/29.5 = %lf\n", f/29.5);
Working With Large Projects
Large projects may potentially involve many hundreds of source files (modules)
Global variables and functions in one module may be accessed in other modules
Global variables and functions may be specifically hidden inside a module
Maintaining consistency between files can be a problem
Data Sharing Example
extern float step;
void print_table(double, float);
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
extern float step;
void print_table(double, float);
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
#include <stdio.h>
float step;
void print_table(double start, float stop){
printf("Celsius\tFarenheit\n");for(;start < stop; start += step)
printf("%.1lf\t%.1lf\n", start, start * 1.8 + 32);
}
#include <stdio.h>
float step;
void print_table(double start, float stop){
printf("Celsius\tFarenheit\n");for(;start < stop; start += step)
printf("%.1lf\t%.1lf\n", start, start * 1.8 + 32);
}
Data Hiding Example
When static is placed before a global variable, or function, the item is locked into the module
static int entries[S_SIZE];static int current;
void push(int value){
entries[current++] = value}
int pop(void){
return entries[--current];}
static void print(void){}
static int entries[S_SIZE];static int current;
void push(int value){
entries[current++] = value}
int pop(void){
return entries[--current];}
static void print(void){}
void push(int value);int pop(void);void print(void);extern int entries[];
int main(void){
push(10); push(15);printf("%i\n", pop());
entries[3] = 77;
print();
return 0;}
void push(int value);int pop(void);void print(void);extern int entries[];
int main(void){
push(10); push(15);printf("%i\n", pop());
entries[3] = 77;
print();
return 0;}
Disaster!
extern float step;
void print_table(double, float);
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
extern float step;
void print_table(double, float);
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
#include <stdio.h>
double step;
void print_table(double start, double stop){
printf("Celsius\tFarenheit\n");for(;start < stop; start += step)
printf("%.1lf\t%.1lf\n", start, start * 1.8 + 32);
}
#include <stdio.h>
double step;
void print_table(double start, double stop){
printf("Celsius\tFarenheit\n");for(;start < stop; start += step)
printf("%.1lf\t%.1lf\n", start, start * 1.8 + 32);
}
Use Header Files
Maintain consistency between modules by using header files
NEVER place an extern declaration in a module NEVER place a prototype of a non static (i.e.
sharable) function in a module
Getting it Right
extern double step;
void print_table(double, double);
extern double step;
void print_table(double, double);
#include "project.h"
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
#include "project.h"
int main(void){
step = 0.15F;
print_table(0.0, 5.5F);
return 0;}
#include <stdio.h>#include "project.h"
double step;
void print_table(double start, double stop){
}
#include <stdio.h>#include "project.h"
double step;
void print_table(double start, double stop){
}
project.h
Be as Lazy as Possible
Get the preprocessor to declare the variables too!
#if defined(MAIN)#define EXTERN#else#define EXTERN extern#endif
EXTERN double step;EXTERN long current;EXTERN short res;
#if defined(MAIN)#define EXTERN#else#define EXTERN extern#endif
EXTERN double step;EXTERN long current;EXTERN short res;
#define MAIN#include "globals.h"
#define MAIN#include "globals.h"
#include "globals.h"#include "globals.h" #include "globals.h"#include "globals.h"
main.cfirst.c second.c
Summary
A union may store values of different types at different times
enum provides an automated way of setting up constants
The preprocessor allows constants and macros to be created
Data and functions may be shared between modules
static stops sharing of data and functions Use the preprocessor in large, multi module
projects
C and the Heap
What is the Heap? Dynamic arrays The calloc/malloc/realloc and free routines Dynamic arrays of arrays Dynamic data structures
What is the Heap?
An executing program is divided into four parts: Stack: provides storage for local variables, alters
size as the program executes Data segment: global variables and strings stored
here. Fixed size. Code segment: functions main, printf, scanf
etc. stored here. Read only. Fixed size Heap: otherwise known as “dynamic memory”
the heap is available for us to use and may alter size as the program executes
How Much Memory?
With simple operating systems like MS-DOS there may only be around 64k available (depending on memory model and extended memory device drivers)
With complex operating systems using virtual memory like Unix, NT, OS/2, etc. it can be much larger, e.g. 2GB
In the future (or now with NT on the DEC Alpha) this will be a very large amount (17 thousand million GB)
Dynamic Arrays
Arrays in C have a fundamental problem - their size must be fixed when the program is written
There is no way to increase (or decrease) the size of an array once the program is compiled
Dynamic arrays are different, their size is fixed at run time and may be changed as often as required
Only a pointer is required
Using Dynamic Arrays
The following steps create a dynamic array: Declare a pointer corresponding to the desired
type of the array elements Initialise the pointer via calloc or malloc using
the total storage required for all the elements of the array
Check the pointer against NULL Increase or decrease the number of elements by
calling the realloc function Release the storage by calling free
calloc/malloc Example
#include <stdio.h>#include <stdlib.h>
int main(void){
unsigned i, s;double *p;
printf("How many doubles? ");scanf("%u", &s);
if((p = calloc(s, sizeof(double))) == NULL) {fprintf(stderr, "Cannot allocate %u bytes "
"for %u doubles\n", s * sizeof(double), s);return 1;
}for(i = 0; i < s; i++)
p[i] = i;
free(p);
return 0;}
#include <stdio.h>#include <stdlib.h>
int main(void){
unsigned i, s;double *p;
printf("How many doubles? ");scanf("%u", &s);
if((p = calloc(s, sizeof(double))) == NULL) {fprintf(stderr, "Cannot allocate %u bytes "
"for %u doubles\n", s * sizeof(double), s);return 1;
}for(i = 0; i < s; i++)
p[i] = i;
free(p);
return 0;}
here we access the “s” doubles from 0..s-1
if((p = malloc(s * sizeof(double))) == NULL) {if((p = malloc(s * sizeof(double))) == NULL) {
all of the allocated memory is freed
realloc Exampledouble *p;double *p2;
if((p = calloc(s, sizeof(double))) == NULL) {fprintf(stderr, "Cannot allocate %u bytes "
"for %u doubles\n", s * sizeof(double), s);return 1;
}
printf("%u doubles currently, how many now? ", s);scanf("%u", &s);
p2 = realloc(p, s * sizeof(double));
if(p2 == NULL) {fprintf(stderr, "Could not increase/decrease array "
"to contain %u doubles\n", s);free(p);return 1;
}p = p2;
free(p);
double *p;double *p2;
if((p = calloc(s, sizeof(double))) == NULL) {fprintf(stderr, "Cannot allocate %u bytes "
"for %u doubles\n", s * sizeof(double), s);return 1;
}
printf("%u doubles currently, how many now? ", s);scanf("%u", &s);
p2 = realloc(p, s * sizeof(double));
if(p2 == NULL) {fprintf(stderr, "Could not increase/decrease array "
"to contain %u doubles\n", s);free(p);return 1;
}p = p2;
free(p);
calculate new array size and allocate
storage
pointer “p” is invalid at this point, so a new value is assigned to it
pointer “p” is still valid at this point
realloc can do it all
The routines malloc and free are almost redundant since realloc can do it all
There is some merit in calloc since the memory it allocates is cleared to zero
p = malloc(s * sizeof(double));p = malloc(s * sizeof(double));
p = realloc(NULL, s * sizeof(double));p = realloc(NULL, s * sizeof(double));
free(p);free(p);
realloc(p, 0);realloc(p, 0);
Allocating Arrays of Arrays
Care must be taken over the type of the pointer used when dealing with arrays of arrays
float *p;
p = calloc(s, sizeof(float));
float *p;
p = calloc(s, sizeof(float));
float **rain;
rain = calloc(s, 365 * sizeof(float));
float **rain;
rain = calloc(s, 365 * sizeof(float));
float (*rainfall)[365];
rainfall = calloc(s, 365 * sizeof(float));
rainfall[s-1][18] = 4.3F;
float (*rainfall)[365];
rainfall = calloc(s, 365 * sizeof(float));
rainfall[s-1][18] = 4.3F;
Dynamic Data Structures
It is possible to allocate structures in dynamic memory too
struct Node {int data; struct Node *next_in_line;
};
struct Node* new_node(int value){
struct Node* p;
if((p = malloc(sizeof(struct Node))) == NULL) {fprintf(stderr, "ran out of dynamic memory\n");exit(9);
}p->data = value; p->next_in_line = NULL;
return p;}
struct Node {int data; struct Node *next_in_line;
};
struct Node* new_node(int value){
struct Node* p;
if((p = malloc(sizeof(struct Node))) == NULL) {fprintf(stderr, "ran out of dynamic memory\n");exit(9);
}p->data = value; p->next_in_line = NULL;
return p;}
Linking the List
struct Node *first_node, *second_node, *third_node, *current;
first_node = new_node(-100);
second_node = new_node(0);
first_node->next_in_line = second_node;
third_node = new_node(10);
second_node->next_in_line = third_node;
current = first_node;while(current != NULL) {
printf("%i\n", current->data);current = current->next_in_line;
}
struct Node *first_node, *second_node, *third_node, *current;
first_node = new_node(-100);
second_node = new_node(0);
first_node->next_in_line = second_node;
third_node = new_node(10);
second_node->next_in_line = third_node;
current = first_node;while(current != NULL) {
printf("%i\n", current->data);current = current->next_in_line;
}
Summary
The heap and stack grow towards one another Potentially a large amount of heap storage is
available given the right operating system The routines malloc, calloc, realloc and free
manipulate heap storage Only realloc is really necessary Allocating dynamic arrays Allocating dynamic arrays of arrays Allocating dynamic structures